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3 years ago

Use the half-reactions of the reaction Au(OH)3 + HI -> Au +I2 +H2O to answer the questions

Chemistry

2 answers:

kirill115 [55]3 years ago

4 0

Enter an equation of a chemical reaction and click 'Balance'. The answer will appear below
Always use the upper case for the first character in the element name and the lower case for the second character. Examples: Fe, Au, Co, Br, C, O, N, F. Compare: Co - cobalt and CO - carbon monoxide
To enter an electron into a chemical equation use {-} or e
To enter an ion specify charge after the compound in curly brackets: {+3} or {3+} or {3}.
Example: Fe{3+} + I{-} = Fe{2+} + I2
Substitute immutable groups in chemical compounds to avoid ambiguity.
For instance equation C6H5C2H5 + O2 = C6H5OH + CO2 + H2O will not be balanced,
but PhC2H5 + O2 = PhOH + CO2 + H2O will
Compound states [like (s) (aq) or (g)] are not required.
If you do not know what products are enter reagents only and click 'Balance'. In many cases a complete equation will be suggested.

kramer3 years ago

3 0

Answer: gold gains 3

Iodine loses 1

The total is 6

Explanation:

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xenn [34]

Answer:I believe it is 5

Explanation:

8 0

3 years ago

Read 2 more answers

What volume. In liters, of H2O(g) measured at STP is produced by the combustion of 15.63 g of natural gas (CH4) according to the

fomenos

Answer:

V = 43.95 L

Explanation:

Given data:

Mass of CH₄ decomposed = 15.63 g

Volume of H₂O produced at STP = ?

Solution:

Chemical equation:

CH₄ + 2O₂ → 2H₂O + CO₂

Number of moles of CH₄:

Number of moles = mass/molar mass

Number of moles = 15.63 g/ 16 g/mol

Number of moles = 0.98 mol

Now we will compare the moles of H₂O with CH₄.

CH₄ : H₂O

1 : 2

0.98 : 2×0.98 = 1.96 mol

Volume of hydrogen:

PV = nRT

1 atm × V = 1.96 mol × 0.0821 atm.L/mol.K × 273.15 K

V = 43.95atm.L / 1atm

V = 43.95 L

3 0

2 years ago

in a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

dimulka [17.4K]

This is an incomplete question, here is a complete question.

Nitroglycerine (C₃H₅N₃O₉) explodes with tremendous force due to the numerous gaseous products. The equation for the explosion of Nitroglycerine is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

A scientist conducts an experiment to characterize a bomb containing nitroglycerine. She uses a steel, ridge container for the test.

Volume of rigid steel container: 1.00 L

Molar mass of Nitroglycerine: 227 g/mol

Temperature: 300 K

Amount of Nitroglycerine tested: 227 g

Value for ideal gas constant, R: 0.0821 L.atm/mol.K

In a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answer : The partial pressure of the water vapor is, 20.01 atm

Explanation :

First we have to calculate the moles of $C_3H_5N_3O_9$

$\text{Moles of }C_3H_5N_3O_9=\frac{\text{Given mass }C_3H_5N_3O_9}{\text{Molar mass }C_3H_5N_3O_9}=\frac{227g}{227g/mol}=1mol$

Now we have to calculate the moles of $CO_2,O_2,N_2\text{ and }H_2O$

The balanced chemical reaction is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

From the balanced chemical reaction we conclude that,

As, 4 moles of $C_3H_5N_3O_9$ react to give 12 moles of $CO_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{12}{4}=3$ moles of $CO_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 1 moles of $O_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{1}{4}=0.25$ moles of $O_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 6 moles of $N_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{6}{4}=1.5$ moles of $N_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 10 moles of $H_2O$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{10}{4}=2.5$ moles of $H_2O$

Now we have to calculate the mole fraction of water.

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CO_2+\text{Moles of }O_2+\text{Moles of }N_2}$

$\text{Mole fraction of }H_2O=\frac{2.5}{2.5+3+0.25+1.5}=0.345$

Now we have to calculate the partial pressure of the water vapor.

According to the Raoult's law,

$p_{H_2O}=X_{H_2O}\times p_T$

where,

$p_{H_2O}$ = partial pressure of water vapor gas = ?

$p_T$ = total pressure of gas = 58 atm

$X_{H_2O}$ = mole fraction of water vapor gas = 0.345

Now put all the given values in the above formula, we get:

$p_{H_2O}=X_{H_2O}\times p_T$

$p_{H_2O}=0.345\times 58atm=20.01atm$

Therefore, the partial pressure of the water vapor is, 20.01 atm

3 0

3 years ago

If a proton and electron were 0.40nm apart, predict what the force of attraction might be

Aleks [24]

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3 0

3 years ago

What is the boiling point of 0.464 m lactose in water? (Kb of water = 0.512 oC/m). Enter your answer to 3 decimal places.

kirill [66]

Answer:

Boiling point for the solution is 100.237°C

Explanation:

We must apply colligative property of boiling point elevation

T° boiling solution - T° boiling pure solvent = Kb . m

m = molalilty (a given data)

Kb = Ebulloscopic constant (a given data)

We know that water boils at 100°C so let's replace the information in the formula.

T° boiling solution - 100°C = 0.512 °C/m . 0.464 m

T° boiliing solution = 0.512 °C/m . 0.464 m + 100°C → 100.237 °C

3 0

3 years ago