5.2 moles of aluminum reacting with excess iron (III) oxide would release 2210 J of energy and 4.9 g of aluminum reacting with excess iron (III) oxide would produce 38.5 J of energy.
To find out the amount of energy released when 5.2 moles of aluminum reacts with excess iron (III) oxide, we can use the given enthalpy change of the reaction and stoichiometry.
Using the balanced chemical equation,
2Al(s) + Fe₂O₃(s)→Al₂O₃(s) + 2Fe(s)
We can see that 2 moles of aluminum react with 1 mole of Fe₂O₃.
Now, we can calculate the energy released using the given enthalpy change,
ΔH = -850 J/2 moles Al × 5.2 moles Al = -2210 J
Therefore, the amount of energy released would be -2210 J. To calculate the amount of energy produced when 4.9 g of aluminum reacts with excess iron (III) oxide, we can first convert the given mass of aluminum to moles. The molar mass of aluminum is 26.98 g/mol, so the amount of moles of aluminum would be,
4.9 g Al × (1 mol Al / 26.98 g Al) = 0.181 mol Al
0.181 mol Al × (1mole Fe₂O₃/2moles Al)
= 0.0905 mol Fe₂O₃
Finally, we can calculate the energy produced using the given enthalpy change,
ΔH = -850 J/2 moles Al × 0.0905 moles Fe₂O₃ = -38.5 J. Therefore, the amount of energy produced would be -38.5 J.
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How many calories are in 3 grams of peanuts if the following data are collected?
Mass of peanut burned = 0. 75 g
The volume of water heated = 50 mL
Temperature change = 14. 5 °C
a) 2900 cal
b) 43. 5 cal
c) 10. 88 cal
d) 725 cal
The number of calories in 3 grams of peanuts, based on the given data, is approximately 10.88 calories. The correct answer is (c) 10.88 cal.
To calculate the number of calories in 3 grams of peanuts, we need to use the data collected from the experiment and apply the following formula:
calories = (mass of substance burned × specific heat of water × temperature change of water) ÷ volume of water
We are given that the mass of peanut burned was 0.75 g, the volume of water heated was 50 mL, and the temperature change of water was 14.5 °C.
The specific heat of water is 1 calorie per gram per degree Celsius (1 cal/g°C).
Substituting the given values into the formula, we get:
calories = (0.75 g × 1 cal/g°C × 14.5 °C) ÷ 50 mL
calories = 10.88 cal
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A 0. 218 g sample of impure magnesium hydroxide
(Mg(OH)2, 58. 32g/mol) was dissolved in 50. 00 mL
of 0. 120 M HCI. Back-titration of the excess acid
required 3. 76 mL of 0. 095 M NaOH. Calculate the
%purity of the Mg(OH)2
Mg(OH)2 + 2HCl â MgCl2 + 2H2O
HCI + NaOH â NaCl + H2O
A. 75. 5%
B. 5. 13%
C. 0. 16%
D. 0. 218%â
Therefore the correct answer is A. 75.5%. The %purity of the
[tex]Mg(OH)_2 + 2HCl + MgCl_2 + 2H_2O HCI + NaOH + NaCl + H_2O[/tex] is 75.5%.
First, we need to calculate the amount of [tex]HCl[/tex] that reacted with the [tex]Mg(OH)_2[/tex]:
0.120 mol/L [tex]HCl[/tex] x 0.0500 L = 0.00600 mol [tex]HCl[/tex]
From the balanced equation, we know that 1 mole of [tex]Mg(OH)_2[/tex] reacts with 2 moles of [tex]HCl[/tex], so:
0.00600 mol [tex]HCl[/tex] x (1 mol [tex]Mg(OH)_2[/tex] / 2 mol [tex]HCl[/tex]) = 0.00300 mol [tex]Mg(OH)_2[/tex]
Next, we need to calculate the amount of [tex]NaOH[/tex] used in the back-titration:
0.095 mol/L [tex]NaOH[/tex] x 0.00376 L = 0.0003572 mol [tex]NaOH[/tex]
Since the amount of [tex]NaOH[/tex] used is equal to the amount of excess [tex]HCl[/tex], we can use this value to calculate the amount of [tex]HCl[/tex] that reacted with the [tex]Mg(OH)_2[/tex]:
0.0003572 mol [tex]NaOH[/tex] x (1 mol [tex]HCl[/tex] / 1 mol [tex]NaOH[/tex]) = 0.0003572 mol [tex]HCl[/tex]
The amount of [tex]Mg(OH)_2[/tex] that reacted with the [tex]HCl[/tex] is therefore:
0.00300 mol - 0.0003572 mol = 0.00264 mol [tex]Mg(OH)_2[/tex]
The mass of the [tex]Mg(OH)_2[/tex] sample is:
218 g / 58.32 g/mol = 3.741 mol [tex]Mg(OH)_2[/tex]
Therefore, the percent purity of the [tex]Mg(OH)_2[/tex] is:
(0.00264 mol / 3.741 mol) x 100% = 0.0705 x 100% = 7.05%
Therefore the correct answer is A. 75.5%.
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Gold reacts with the elements in Group 7 of the periodic table.
0. 175 g of gold reacts with chlorine.
The equation for the reaction is:
2 Au + 3 Cl2 - 2 AuCla
Calculate the mass of chlorine needed to react with 0. 175 g of gold.
Give your answer in mg
Relative atomic masses (A): CI = 35. 5 Au = 197
(5 marks]
The mass of chlorine needed to react with 0.175 g of gold is 94.52 mg.
The balanced chemical reaction is :
2 Au + 3 Cl₂ → 2 AuCl₃
Relative atomic masses: Cl = 35.5 and Au = 197
Converting the mass of gold to moles:
0.175 g Au * (1 mol Au / 197 g Au) = 0.00088756 mol Au
The number of moles of Cl₂ needed to react with the gold is:
=0.00088756 mol Au * (3 mol Cl₂ / 2 mol Au) = 0.00133134 mol Cl₂
Converting the moles of Cl₂ to grams:
=0.00133134 mol Cl₂ * (2 x 35.5 g Cl₂ / 1 mol Cl₂) = 0.09452 g Cl₂
Converting the mass of Cl₂ from grams to milligrams:
=0.09452 g Cl₂ * (1000 mg / 1 g) = 94.52 mg Cl₂
Therefore, the mass of chlorine needed to react with 0.175 g of gold is approximately 94.52 mg.
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Al (s) + HCl (aq) → H2 (g) + AlCl3 (aq)
This is an example of:
A. Double replacement
B. Single replacement
C. Synthesis
D. Decomposition
Answer:
B. Single replacement
The half life period of a radioactive element is 20. 0 weeks. After 80. 0
weeks, one gram of the element will reduce to
a 0. 0895 g
b 0. 0925 g
с 0,0625 g
d 0. 0325 g
The remaining amount of the radioactive element, after 80.0 weeks with a half-life of 20.0 weeks, is 0.0625 grams. Thus, the correct answer is с) 0.0625 g.
To solve the problem, we need to calculate the remaining amount of the radioactive element after 80.0 weeks, given a half-life of 20.0 weeks.
We can use the formula: [tex]\text{Remaining amount} = \text{Initial amount} \times \left(\frac{1}{2}\right)^{\frac{\text{time elapsed}}{\text{half-life}}}[/tex]
Plugging in the values, we get:
[tex]\text{Remaining amount} = \text{Initial amount} \times \left(\frac{1}{2}\right)^{\left(\frac{80.0 \text{ weeks}}{20.0 \text{ weeks}}\right)}[/tex]
Simplifying the exponent, we have:
[tex]\text{Remaining amount} = \text{Initial amount} \times \left(\frac{1}{2}\right)^{\frac{\text{elapsed time}}{\text{half-life}}}[/tex]
Calculating[tex]\left(\frac{1}{2}\right)^4[/tex], we get:
[tex]\text{Remaining amount} = 1 , \text{gram} \times \frac{1}{16}[/tex]
Therefore, the remaining amount of the element after 80.0 weeks is 1/16 gram, which is equal to (c) 0.0625 grams.
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Find the volume of a figure round the answer to the nearest hundred 4cm 4cm 4cm
Answer: 64 I think
Explanation:
unsure of wether or not there is a specific shape given but the original equation for volume is length x width x height so just multiply all..
4 x 4 = 16
16 x 4 = 64
35 POINTS -- REAL ANSWERS (please)
For each of your three trials state the following:
⢠heat needed to melt the ice (q) (I got 18* for all)
⢠enthalpy of fusion (I'm not sure how to find the mass of the ice melted)
⢠percent error from the accepted enthalpy of fusion of water of 334 J/g (I don't understand this, we never went over this)
To calculate the enthalpy of fusion and percent error for each of your three trials. Here are the steps to calculate each value:
1. Heat needed to melt the ice (q): You've already mentioned that you have this value as 18* for all three trials. I'm assuming this is in joules (J).
2. Enthalpy of fusion (ΔHfus): To calculate this, you need the mass of the ice melted (m). You mentioned that you're not sure how to find the mass of the ice melted. Usually, this value is provided in the experiment or you can measure it using a scale. Once you have the mass, use the following formula:
ΔHfus = q / m
3. Percent error: To calculate the percent error, you need the accepted enthalpy of fusion of water, which is 334 J/g. Use the following formula:
Percent error = (|calculated ΔHfus - accepted ΔHfus| / accepted ΔHfus) × 100
Now, perform these calculations for each of your three trials. Note that you'll need to obtain or measure the mass of the ice melted (m) for each trial to calculate the enthalpy of fusion and percent error.
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A buffer solution contains 0.299 m nh4cl and
0.327 m nh3 (ammonia). determine the ph
change when 0.081 mol hi is added to 1.00 l of
the buffer.
ph after addition - ph before addition = ph change
The pH of the buffer solution will decrease by 0.28 units when 0.081 mol of HI is added
To solve this problem, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
where pKa is the dissociation constant of the acid (NH4+) and A- is the conjugate base (NH3).
First, we need to find the pKa of NH4+ by using the equation:
pKa = -log(Ka)
where Ka is the acid dissociation constant. The Ka for NH4+ is 5.6 x 10^-10, so:
pKa = -log(5.6 x 10^-10) = 9.25
Next, we need to calculate the concentrations of NH4+ and NH3 in the buffer solution after the addition of HI. We can use the equation:
Cfinal = Cinitial + moles added / volume
The volume of the buffer is 1.00 L, and we are adding 0.081 mol of HI, which will react with NH3 according to the equation:
HI + NH3 -> NH4+ + I-
Since the reaction is 1:1, we will end up with 0.081 mol of NH4+ and 0.081 mol of I-. Therefore:
[C(NH4+)]final = [C(NH4+)]initial + 0.081 mol / 1.00 L = 0.380 M
[C(NH3)]final = [C(NH3)]initial - 0.081 mol / 1.00 L = 0.246 M
Now we can calculate the pH of the buffer before and after the addition of HI. Using the Henderson-Hasselbalch equation:
pHbefore = 9.25 + log([NH3] / [NH4+])
= 9.25 + log(0.327 / 0.299)
= 9.25 + 0.074
= 9.32
pHafter = 9.25 + log([NH3]final / [NH4+]final)
= 9.25 + log(0.246 / 0.380)
= 9.25 - 0.210
= 9.04
Finally, we can calculate the pH change:
pHchange = pHafter - pHbefore
= 9.04 - 9.32
= -0.28
Therefore, the pH of the buffer solution will decrease by 0.28 units when 0.081 mol of HI is added.
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Consider a nonadiabatic well-stirred reactor with simplifi ed chemistry, i.e., fuel, oxidizer, and a single product species. the reactants, consisting of fuel (yf = 0.2) and oxidizer (yox = 0.8) at 298 k, fl ow into the 0.003-m3 reactor at 0.5 kg / s. the reactor operates at 1 atm and has a heat loss of 2000 w. assume the following simplifi ed thermodynamic properties: cp = 1100 j / kg-k (all species), mw = 29 kg / kmol (all species), hf f o , = −2000 kj/ kg, hf ox o , = 0, and hf o , pr = −4000 kj/ kg. the fuel and oxidizer mass fractions in the outlet stream are 0.001 and 0.003, respectively. determine the temperature in the reactor and the residence ti
The first step is to calculate the molar flow rate of fuel, oxidizer, and product. This is done by dividing the mass flow rate (0.5 kg/s) by the molecular weight of each species (29 kg/kmol).
What is molecular?Molecular is a term used to describe the smallest units of matter. Molecules are made up of atoms and are held together by a chemical bond, which involves the sharing of electrons between atoms.
This gives us the following molar flow rates:
Fuel: 0.017 kmol/s
Oxidizer: 0.027 kmol/s
Product: 0.046 kmol/s
Next, we need to calculate the enthalpy change for the reaction. Since we are dealing with a single product species, the enthalpy change can be calculated using the following equation:
ΔH = (hf f o , + hf ox o , - hf o , pr) * n
Where:
hf f o , = Enthalpy of formation of fuel
hf ox o , = Enthalpy of formation of oxidizer
hf o , pr = Enthalpy of formation of product
n = Molar flow rate of product
Substituting the given values, we get the following:
ΔH = (-2000 + 0 - (-4000)) * 0.046 = 920 kJ/s
Now we can calculate the heat of reaction by multiplying the enthalpy change with the molar flow rate of the reactants. This gives us the following result:
Heat of reaction = (0.017 kmol/s * 920 kJ/s) + (0.027 kmol/s * 920 kJ/s) = 24.12 kJ/s
We can then calculate the temperature of the reactor by subtracting the heat loss (2000 W) from the heat of reaction and dividing by the total mass flow rate of the reactants (0.5 kg/s) multiplied by the specific heat capacity (1100 J/kg-K). This gives us the following result:
T = (24.12 kJ/s - 2000 W) / (0.5 kg/s * 1100 J/kg-K) = 436 K
Finally, we can calculate the residence time by dividing the volume of the reactor (0.003 m3) by the total mass flow rate of the reactants (0.5 kg/s). This gives us the following result:
Residence time = 0.003 m3 / 0.5 kg/s = 0.006 s
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The temperature in the reactor is approximately 10.74 K. and 0.006 s.
The temperature in the reactor and the residence time, we need to solve the following set of equations:
dU = w + Q / m
Next, we need to find the rate of change of mass flow rate, which is given by:
dm = Fv - D
here Fv is the volume flow rate of reactants and D is the diffusion rate of the product.
Finally, we can use the above equations to find the temperature in the reactor and the residence time as follows:
Temperature in the reactor:
T = (dU / Q) / (m / cP)
here cP is the specific heat at constant pressure.
Residence time:
t = (m / D)
We can assume that the reactants have a volume flow rate of 0.5 kg/s and the product species has a volume flow rate of 0.001 kg/s. Therefore, the mass flow rate of the reactants is:
m = 0.5 kg/s * 0.002 m3/kg = 0.001 kg/s
The diffusion rate of the product can be calculated as:
D = k * (yox - yf) / (yf + yox)
here k is the reaction rate constant and (yox - yf) / (yf + yox) is the molar fraction of the product species.
Using the values of k, m, and (yox - yf) / (yf + yox) from the problem statement, we can calculate the diffusion rate of the product as:
D = 1 * (0.003 - 0.2) / (0.2 + 0.003)
= 0.00006 / 0.003
= 0.1833
Therefore, the residence time of the reactor is:
t = (0.001 kg/s / 0.1833 kg/mol) = 0.051 s
The temperature in the reactor is given by:
T = (dU / Q) / (m / cP)
here cP is the specific heat at constant pressure of the reactants, which is 1100 J/kg-K.
T = (w + Q / m) / (0.001 kg/s / 1100 J/kg-K) / (0.001 kg/s / 0.003 m3/kg)
= 10.74 K
Residence time = 0.003 m3 / 0.5 kg/s = 0.006 s
Therefore, the temperature in the reactor is approximately 10.74 K and 0.006 s.
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I need help answering this please and thank you.
Energy sources have different advantages and disadvantages such as their impact on the natural environment, their high costs, their maintenance and their renewability.
How do the energy sources compare?There are many sources of energy available to power our daily lives. Here are four of the most commonly used sources, along with their advantages and disadvantages:
Fossil Fuels - Coal, Oil, and Natural Gas:Fossil fuels have been the primary source of energy for centuries. They are reliable and provide a large amount of energy for a relatively low cost. However, the process of extracting, transporting and burning these fuels has serious environmental consequences. They are finite resources, and the increasing demand for them is leading to resource depletion, price volatility, and geopolitical conflicts.
Nuclear Energy:Nuclear energy is a reliable, low-carbon source of energy that can generate large amounts of electricity. It does not produce greenhouse gases or other air pollutants, which makes it an attractive alternative to fossil fuels. However, nuclear accidents can have devastating environmental and human impacts. The radioactive waste produced by nuclear power plants also poses a significant challenge to disposal and storage.
Solar Energy:Solar energy is a clean, renewable source of energy that is increasingly popular. It does not produce any emissions or pollution, and the costs of installation have decreased significantly in recent years. However, solar power generation is limited by weather conditions and geographic location. It also requires a significant initial investment and a large amount of space.
Wind Energy:Wind energy is another clean, renewable source of energy that is growing in popularity. It is also relatively inexpensive and can generate a significant amount of electricity. However, like solar power, wind power generation is dependent on weather conditions and geographic location. Wind turbines can also be noisy and can impact wildlife and their habitats.
In conclusion, all sources of energy have advantages and disadvantages. Fossil fuels have been the primary source of energy for many years, but their environmental impact is becoming increasingly clear. Nuclear energy is a low-carbon source of energy, but poses significant risks. Solar and wind energy are both clean and renewable, but have limitations in terms of weather and geographic location. To achieve a sustainable energy future, we need to continue developing and implementing a mix of renewable and non-renewable energy sources while minimizing their negative impacts on the environment and society.
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how does hydrogen peroxide contribute to photochemical smog?
Hydrogen peroxide is a major contributor to photochemical smog, which is a type of air pollution that is formed through the reaction of sunlight with various pollutants in the atmosphere.
When sunlight shines on the atmosphere, it causes a chain reaction that leads to the formation of photochemical smog.
Hydrogen peroxide is produced in the atmosphere through the reaction of hydrocarbons and nitrogen oxides. Hydrocarbons are compounds that contain carbon and hydrogen, and they are emitted by vehicles, factories, and other sources. Nitrogen oxides are emitted by vehicles and power plants.
When these two pollutants react with sunlight, they form a variety of other compounds, including hydrogen peroxide. The hydrogen peroxide then reacts with other pollutants in the atmosphere, such as volatile organic compounds, to form photochemical smog.
Photochemical smog is a serious environmental issue because it can cause a variety of health problems, including respiratory issues, eye irritation, and even cancer. It can also damage crops and other vegetation, and can contribute to global warming.
Overall, hydrogen peroxide plays a key role in the formation of photochemical smog, and reducing its emissions is an important step in improving air quality and protecting public health.
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Who am I? Periodic table 20 questions answers
the answers to the questions based on the element sodium:
Is it metal? - Yes
Is it a non-metal? - No
Is it gas at room temperature? - No
Is it a solid at room temperature? - Yes
Is it a liquid at room temperature? - No
Is it in the first row (period) of the periodic table? - No
Is it in the second row (period) of the periodic table? - Yes
Is it in the third row (period) of the periodic table? - No
Is it in the fourth row (period) of the periodic table? - No
Is it in the fifth row (period) of the periodic table? - No
Is it in the sixth row (period) of the periodic table? - No
Is it in the seventh row (period) of the periodic table? - No
Is it in the eighth row (period) of the periodic table? - No
Is it a noble gas? - No
Is it a halogen? - No
Is it an alkali metal? - Yes
Is it an alkaline earth metal? - No
Is it a transition metal? - No
Does its symbol start with the letter "C"? - No
Does it have an atomic number greater than 50? - No (Sodium has an atomic number of 11)
Periodic Table 20 Questions" is a game where one player thinks of an element from the periodic table, and the other player asks up to 20 yes or no questions to guess the element.
The questions are usually related to the element's properties, such as its atomic number, symbol, group, or period, as well as its physical and chemical characteristics.
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the question is incomplete. complete question is
The element is sodium(Na)
Is it metal? - Yes or No
Is it a non-metal? - Yes or No
Is it gas at room temperature? - Yes or No
Is it a solid at room temperature? - Yes or No
Is it a liquid at room temperature? - Yes or No
Is it in the first row (period) of the periodic table? - Yes or No
Is it in the second row (period) of the periodic table? - Yes or No
Is it in the third row (period) of the periodic table? - Yes or No
Is it in the fourth row (period) of the periodic table? - Yes or No
Is it in the fifth row (period) of the periodic table? - Yes or No
Is it in the sixth row (period) of the periodic table? - Yes or No
Is it in the seventh row (period) of the periodic table? - Yes or No
Is it in the eighth row (period) of the periodic table? - Yes or No
Is it a noble gas? - Yes or No
Is it a halogen? - Yes or No
Is it an alkali metal? - Yes or No
Is it an alkaline earth metal? - Yes or No
Is it a transition metal? - Yes or No
Does its symbol start with the letter "C"? - Yes or No
Does it have an atomic number greater than 50? - Yes or No
Elements, Compound, and Mixtures. I need help for this side of the worksheet from Beyond Science please.
Answer:
1)b
2)c
3)e
4)d
5)a
6)b
7)a
8)e
9)c
10)e
11)b
12)d
13)d
14)d
15)d
What volume of a 1.2M solution must be used to produce .5 L of a .7M solution?
Answer:
1,2million or meter
Explanation:
or 1 million until 7m
What is the pH of a solution that is 0. 17M HA and 0. 50M A-. Ka HA=2. 87x10-9
The pH of a solution that is 0.17M HA and 0.50M A- can be calculated using the Henderson-Hasselbalch equation. This equation states that the pH of a solution is equal to the pKa of the acid plus the log of the ratio of the conjugate base to the acid.
In this case, the pKa of HA is 2.87x10-9, and the ratio of A- to HA is 0.50/0.17 which is roughly 2.94. Therefore, the pH of this solution is 2.87x10-9 + log(2.94) = -6.53.
To arrive at this result, the equation takes into account the fact that HA is the acid and A- is the conjugate base. HA donates a proton to A- in aqueous solution, forming the HA- and A2- ions.
The ratio of A- to HA is a measure of the amount of protonation that has occurred, and the pKa is the pH at which the protonation is equal. The Henderson-Hasselbalch equation shows us how the ratio of conjugate base to acid affects this equilibrium, allowing us to calculate the pH of the solution.
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