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3 years ago

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How does temperature relate to density? A- As temperature increases density increases. B- As temperature decreases density decre

ases. C- As temperature decreases density increases. D- Temperature does not change density.

1 answer:

SVEN [57.7K]3 years ago

8 0

Answer:

The answer is c

Explanation:

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Chrissy is doing her chemistry homework and comes across the chemical formula for 3 molecules of nitric acid, represented by 3HN

dybincka [34]

Answer:

C

Explanation:

The 3 goes to every term in the molecule

3 NHO3

So its

3x1 N's

3x1 H's

3x3 O's

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Atoms that make up a compound are bonded together; they cannot be separated by

Tpy6a [65]

Answer:

true

Explanation:

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How do you calculate the number of photons having a wavelength of 10.0 micrometers required to produce 1.0 kilojoules of energy

shtirl [24]

To calculate this,

We know that energy is 1 photon
E = hc/wavelenth
wavelength of 10.0 m

Solution:
h = 6.626 x 10^-34 Jsec
C = 2.9979 x 10^8 m/sec
E = 6.626 10^-34 * 2.9979 10^8 / 10 = 1.9864 10^-26J

Then, the number of photons is computed by:

n = 1000 / 1.9864 10^-26 = 5.04 10^28 photons

4 0

3 years ago

An alkaline battery produces electrical energy according to the following equation.

Gnesinka [82]

Answer:

Part a: limiting reactant MnO₂

Part b: 12.43 g of Zn(OH)₂

Explanation:

Part a

Data given:

mass of Zn = 37.0 g

mass of MnO₂ = 21.5 g

Limiting reactant = ?

Given Reaction

Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

To determine the limiting reactant first we will look at the reaction

Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

1 mol 2 mol

Convert moles to mass

molar mass of Zn = 65.4 g/mol

Mass of MnO₂ = 55 + 2 (16) = 55 + 32 = 87 g/mol

So,

Zn(s) + 2 MnO₂(s) + H₂O(l) ----------> Zn(OH)₂(s) + Mn₂O₃(s)

1 mol (65.4 g/mol) 2 mol (87 g/mol)

65 g 174 g

So its clear from the reaction that 65 g Zn react with 174 g of MnO₂.

now if we look at the given amounts the amount MnO₂ is less then the amount of Zn but in actual calculation amount of MnO₂ is more then amount of zinc.

So, for MnO₂ if we calculate the needed amount of zinc

So apply unity formula

65 g Zn react ≅ 174 g of MnO₂

X g of Zn ≅ 21.5 g of MnO₂

Do cross multiplication

X g Zn react = 65 g x 21.5 g / 174 g

X g of Zn ≅ 8.032 g

So, 8.032 g of zinc will react out of 37.0 grams. the remaining will be in excess.

So MnO₂ will be consumed completely an it will be limiting reactant.

____________

part b

Data given:

mass of Zn = 37.0 g

mass of MnO₂ = 21.5 g

Mass of Zn(OH)₂ produced = ?

Given Reaction

Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

Solution:

As from the part A we come to know that MnO₂ is limiting reactant, so the amount of Zn(OH)₂ will depend on the amount of MnO₂.

So first we convert mass of MnO₂ to moles

Formula Used

no. of moles = mass in grams / molar mass

Mass of MnO₂ = 55 + 2 (16)

Mass of MnO₂ = 55 + 32 = 87 g/mol

Put values in the above equation

no. of moles = 21.5 g / 87 g/mol

no. of moles = 0.25 moles

Now,

Look at the reaction

Zn(s) + 2 MnO₂(s) + H₂O(l) ---------->Zn(OH)₂(s) + Mn₂O₃(s)

2 mol 1 mol

So its clear from the reaction that 2 mole of MnO₂ gives 1 mole of Zn(OH)₂

then how many moles of Zn(OH)₂ will be produce by 0.25 moles of MnO₂

So.

apply unity formula

2 mol of MnO₂ ≅ 1 mole of Zn(OH)₂

0.25 moles of MnO₂ ≅ X mole of Zn(OH)₂

Do cross multiplication

X mole of Zn(OH)₂ = 1 mole x 0.25 mol x / 2 mol

X mole of Zn(OH)₂ ≅ 0.125

Now Conver moles of Zn(OH)₂ to mass

Formula used

mass in grams = no. of moles x molar mass

Molar mass of Zn(OH)₂

Molar mass of Zn(OH)₂ = 65.4 + 2 (16 + 1)

Molar mass of Zn(OH)₂ = 65.4 + 2 (17)

Molar mass of Zn(OH)₂ = 65.4 + 34 = 99.4 g/mol

Put values in above equation

mass in grams = 0.125 mol x 99.4 g/mol

mass in grams = 12.43 g

So,

12.43 g of Zn(OH)₂ will be produce.

5 0

3 years ago

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

svetlana [45]

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

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

2. Using a gasoline powered vehicle

(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.

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

3. Using an electric vehicle

(a) Theoretical energy used

$\text{Theor. Energy} = \text{347 mi} \times \dfrac{\text{1 kWh}}{\text{5.2 mi}} = \text{66.7 kWh theor.}$

(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}\$

(c) Combustion of CH₄

CH₄ + 2O₂ ⟶ CO₂ +2 H₂O

(d) Equivalent volume of CO₂

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

$V= 2.82\times 10^{5}\text{ kJ} \times \dfrac{\text{1 mol methane}}{\text{802.3 kJ}} \times \dfrac{\text{1 mol CO$_{2}$} }{\text{1 mol methane}} \times \dfrac{ \text{24.0 L}}{ \text{1 mol CO$_{2}$}}\\\\ \times \dfrac{\text{1 gal}}{\text{3.875 L}} = \textbf{2180 gal}$

4. Comparison

$\dfrac{V_{\text{gasoline}}}{V_{\text{electric}}} = \dfrac{10800}{2180} = 5.0\\\\ \text{An gasoline-powered vehicle has }\\ \boxed{\textbf{ five times the carbon footprint}}\text{ of an electric vehicle.}$

6 0

3 years ago