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3 years ago

The balanced chemical equation for the reaction of copper (Cu) and silver nitrate (AgNO3) is shown below.

2 answers:

8 0

Remark
The balance numbers in front of Ag and AgNO3 are both 2. That number is in moles.

Rule: if the moles are the same in the equation, then whatever you are given for one, will be the same for the other. So you have 0.854 moles of Ag. You will also have 0.854 moles of AgNO3

Answer: 0.854 <<<<<

Snezhnost [94]3 years ago

4 0

Answer: Moles of $AgNO_3$ that must be reacted are 0.854 moles.

Explanation:

For the given chemical reaction, the equation follows:

$Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2$

By Stoichiometry of the reaction:

2 moles of silver are produced when 2 moles of silver nitrate is reacted.

So, 0.854 moles of silver is formed when = $\frac{2}{2}\times 0.854=0.854moles$ of silver nitrate is reacted.

Thus, moles of $AgNO_3$ that must be reacted are 0.854 moles.

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3 years ago

A civil engineering student is considering buying an electric car. However, the conscious student wants to make sure her carbon

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Answer:

$\text{An gasoline-powered vehicle has }\\\boxed{\text{five times the carbon footprint}}\text{ of an electric vehicle.}$

Explanation:

1. Miles travelled in an average month

$\text{Miles} = \text{1 mo} \times \dfrac{\text{52 wk}}{\text{12 mo}} \times \dfrac{\text{5 da}}{\text{1 wk}} \times \dfrac{\text{16 mi}}{\text{1 da}} = \text{347 mi}$

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(a) Moles of heptane used

$n = \text{347 mi} \times \dfrac{\text{1 gal}}{\text{36.5 mi}}\times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{679.5 g}}{\text{1L}} \times \dfrac{ \text{1 mol}}{\text{100.20 g}}= \text{244 mol}$

(b) Equation for combustion

C₇H₁₆ + O₂ ⟶ 7CO₂ + 8H₂O

(c) Moles of CO₂ formed

$n = \text{244 mol heptane} \times \dfrac{\text{7 mol CO$_{2}$}}{\text{1 mol heptane}} = \text{1710 mol CO$_{2}$}$

(d) Volume of CO₂ formed

At 20 °C and 1 atm, the molar volume of a gas is 24.0 L.

$V = \text{1710 mol } \times \dfrac{\text{24.0 L}}{\text{1 mol}} \times \dfrac{\text{1 gal}}{\text{3.785 L}} = \textbf{10 800 gal}$

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(b) Actual energy used

The power station is only 85 % efficient.

$\text{Actual energy used} = \text{66.7 kWh theor.} \times \dfrac{\text{100 kWh actual}}{\text{ 85 kWh theor.}}\times \dfrac{\text{3600 kJ}}{\text{1 kWh}}\\\\ = 2.82\times 10^{5} \text{ kJ}\$

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The heat of combustion of methane is -802.3 kJ·mol⁻¹

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4. Comparison

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6 0

3 years ago

What is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa

Nezavi [6.7K]

60.7 ml is the volume of a sample of CO2 at STP that has a volume of 75.0mL at 30.0°C and 91kPa.

Explanation:

Data given:

V1 = 75 ml

T1 = 30 Degrees or 273.15 + 30 = 303.15 K

P1 = 91 KPa

V2 =?

P2 = 1 atm or 101.3 KPa

T2 = 273.15 K

At STP the pressure is 1 atm and the temperature is 273.15 K

applying Gas Law:

$\frac{P1VI}{T1}$ = $\frac{P2V2}{T2}$

putting the values in the equation of Gas Law:

V2 = $\frac{P1V1T2}{P2T1}$

V2 = $\frac{91 X 75 X 273.15}{303.15 X 101.3}$

V2 = 60.7 ml

at STP the volume of carbon dioxide gas is 60.7 ml.

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3 years ago

If you have 40 grams of potassium nitrate in 100 grams of water at 20 C:

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Answer:

Explanation:The final homogenous solution, after cooling it to 40°C, will contain 47 g of potassium sulfate disolved in 150 g of water, so you can calculate the amount disolved per 100 g of water in this way:

[47 g of solute / 150 g of water] * 100 g of g of water = 31.33 grams of solute in 100 g of water.

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To make a saturated solution, 15 grams of potassium sulfate would dissolve in 100 g of water.

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