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3 years ago

What information can a foliated metamorphic rock provide you about the conditions under which it formed?

1 answer:

4 0

A foliated rock forms when the minerals are realigned due to the presence of high temperature and pressure. The minerals align in such a way that the axes are directed to where the pressure was applied.

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Calculate the standard heat of reaction for the following methane-generating reaction of methanogenic bacteria: 4CH3NH2(g) + 2H2

PIT_PIT [208]

Answer: The standard heat for the given reaction is -138.82 kJ

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]$

We are given:

$\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ$

Hence, the standard heat for the given reaction is -138.82 kJ

3 0

3 years ago

A population of large predator

dangina [55]

Answer:

they will eat a lot of fish and make other animals go hungry

Explanation:

logic

3 0

3 years ago

Read 2 more answers

What is the partial pressure of a gas mixture in a mixture of gases?

lesantik [10]

The answer is b. The sum of the individual gas pressure in the mixture

8 0

3 years ago

Read 2 more answers

assume in a different experiment, you prepare a mixture containing 10.0 M FeSCN2+, 1.0 M H+, 0.1 MFe3+ and 0.1 M HSCN. Is the in

Katarina [22]

Answer:

The mixture is not in equilibrium, the reaction will shift to the left.

Explanation:

Based on the equilibrium:

Fe³⁺+ HSCN ⇄ FeSCN²⁺ + H⁺

kc = 30 = [FeSCN²⁺] [H⁺] / [Fe³⁺] [HSCN]

Where [] are concentrations at equilibrium. The reaction is in equilibrium when the ratio of concentrations = kc

Q is the same expression than kc but with [] that are not in equilibrium

Replacing:

Q = [10.0M] [1.0M] / [0.1M] [0.1M]

Q = 1000

As Q > kc, the reaction will shift to the left in order to produce Fe³⁺ and HSCN untill Q = Kc

7 0

3 years ago

Researchers used a combustion method to analyze a compound used as an antiknock additive in gasoline. A 9.394 mg sample of the c

LuckyWell [14K]

Answer:

The percent composition of the compound is 90.5 % C and 9.5 % H

Explanation:

Step 1: Data given

Mass of compound = 9.394 mg

Mass of CO2 yielded = 31.154 mg

Mass of H2O yielded = 7.977 mg

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: Calculate moles CO2

moles of CO2 = (0.031154 g / 44.01 g/mol) = 7.08 * 10^-4 mol CO2

Step 3: Calculate moles C

moles of C = moles of CO2 * (1 mol C / 1 mol CO2)

moles of C = 7.08 * 10^-4 mol

Step 4: Calculate moles H2O

moles of H2O = (0.007977 g / 18.02 g/mol) = 4.43 * 10^-4 mol H2O

Step 5: Calculate moles of H

moles of H = moles of H2O * (2 mol H / 1 mol H2O)

moles of H = 4.43* 10^-4 *2 = 8.86 * 10^-4 mol H

Step 6: Calculate mass of C

mass C = moles C * molar mass C

mass C = 7.08 * 10^-4 mol*12.01 g/mol

mass C = 0.0085 grams

Step 7: Calculate mass of H

mass H = moles H * molar mass H

mass H = 8.86 * 10^-4 mol*1.01 g/mol

mass H = 0.000894 grams

Step 8: Calculate total mass of compound =

0.0085 grams + 0.000894 grams = 0.009394 grams = 9.394 mg

Step 9: Calculate the percent composition:

% C = (8.50 mg / 9.394 mg) x 100 = 90.5%

% H = (0.894 mg / 9.394 mg) x 100 = 9.5%

The percent composition of the compound is 90.5 % C and 9.5 % H

6 0

3 years ago