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s2008m [1.1K]
2 years ago
7

Limestone has a density of 2.72 g/cm3. What is the mass of 8.92 cm3 of limestone?

Chemistry
2 answers:
timurjin [86]2 years ago
8 0
It should be 24.2624 because if there’s 2.72g in every cm3 and there are 8.92 cm3, then you simply multiply the two numbers
hope this helped :)
kati45 [8]2 years ago
3 0

Answer:

\boxed {\boxed {\sf 24.2624 \ g}}}

Explanation:

Density measures mass per volume. The volume is

\rho=\frac{m}{v}

The density of limestone is 2.72 grams per cubic centimeter. We have a piece of limestone with a volume of 8.92 cubic centimeters.

  • ρ= 2.72 g/cm³
  • v= 8.92 cm₃

Substitute the values into the formula.

2.72 \ g/cm^3=\frac{m}{8.92 \ cm^3}

Now we solve for m, the mass, by isolating the variable.

m is being divided by 8.92 cubic centimeters. The inverse of division is multiplication, so we multiply both sides by 8.92 cubic centimeters.

8.92 \ cm^3 * 2.72 \ g/cm^3=\frac{m}{8.92 \ cm^3}* 8.92 \ cm^2

8.92 \ cm^3 * 2.72 \ g/cm^3=m

The units of cubic centimeters cancel out.

24.2624 \ g =m

The mass of this piece of limestone is 24.2624 grams.

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Answer:

-800 kJ/mol

Explanation:

To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).

First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:

Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol

moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄

Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:

ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol

Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).

3 0
2 years ago
Given the following equation, what is the correct form of the conversion factor needed to convert the number of moles O2 to the
jek_recluse [69]

Answer:

Option A says we have 4 moles of Fe for each 3 moles O2

This is correct For 3 moles of O2 consumed, we need 4 moles of Fe to be reacted

Explanation:

Step 1: Data given

Step 2: The balanced equation

4Fe + 3O2 → 2Fe2O3

Step 3: Calculate the mol ratio

For 3 moles O2 we'll have 4 moles Fe

Option A says we have 4 moles of Fe for each 3 moles O2

This is correct For 3 moles of O2 consumed, we need 4 moles of Fe to be reacted

Option b says we have 2 mole Fe2O3 for each 4 moles Fe

This doesnt say anything about O2. So doesn't apply for this question.

Option C says we have 4 moles of Fe for each 2 moles Fe2O3

This is the same as option B, so doesn't apply for this question.

Option D says for each 3 moles of O2 we have 2 Fe2O3

This is true, but doesn't say anything about Fe so doesn't apply here.

4 0
3 years ago
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What two gases have a profound effect on the type of organisms that can live in an area and influences the regions carrying capa
Over [174]

Answer:

Oxygen and Carbon dioxide

Explanation:

Oxygen is required for respiration whereby energy is released from natural occurring nutrients accompanied by the release of water and carbon dioxide. carbon dioxideis also required by plants to photosynthesise.

Oxygen and carbon dioxide in the troposphere supports life as by enabling respiration in organisms and photosynthesise in plants can. Without oxygen in an environment, only life forms that live by anaerobic respiration will thrive. This affects a regions carrying capacity

3 0
3 years ago
Strong bases are good conductors of:
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B. Electricity

Explanation:

8 0
3 years ago
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Helpppooo give answer thanks
DaniilM [7]

Answer:

625 mL

Explanation:

From the question given above, the following data were obtained:

Volume of stock solution (V₁) = 250 mL

Molarity of stock solution (M₁) = 5 M

Molarity of diluted solution (M₂) = 2 M

Volume of diluted solution (V₂) =?

The volume of the diluted solution can be obtained by using the dilution formula as illustrated below:

M₁V₁ = M₂V₂

5 × 250 = 2 × V₂

1250 = 2 × V₂

Divide both side by 2

V₂ = 1250 / 2

V₂ = 625 mL

Therefore, the volume of the diluted solution is 625 mL.

8 0
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