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3 years ago

What two quantities must be known to calculate the density of a sample of matter?

Chemistry

2 answers:

Tanya [424]3 years ago

7 0

The answer is A, mass and volume.

The equation for density, ρ, is ρ = m/V. m is mass and V is volume.

Mkey [24]3 years ago

6 0

Answer:

remember the formula for density:

d = \frac{m}{v}

where d is density, m is mass, and v is volume.

Therefore, B. mass and volume is needed to determine the density

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A sample of limestone (calcium carbonate, CaCO3) is heated at 950 K until it is completely converted to calcium oxide (CaO) and

pav-90 [236]

Answer:

Therefore, volume of CO₂ produced in the first step is 9141.404 L

Explanation:

Equations of reactions:

A: CaCO₃(s) ---> CaO(s) + CO₂(g)

B: CaO(l) + H₂O(l) ---> Ca(OH)₂(s)

Molar mass of CaCO₃ = 100 g; molar mass of CaO = 56 g; molar mass of CO₂ = 44 g molar mass of H₂P = 18 g; molar mass of Ca(OH)₂ = 74 g

From equation B, 1 mole of CaO produces 1 mole of Ca(OH)₂

This means that 56 g of CaO produces 74 g of Ca(OH)₂

mass of CaO that produces 8.47 kg or 8470 g of Ca(OH)₂ = 8470 g * 56/74 = 6409.73 g of CaO

Therefore, 6409.73 g of CaO were produced in reaction A

From reaction A, 1 mole of CaCO₃ produces 1 mole CaO and 1 mole of CO₂

Number of moles of CaO in 6409.73 g = 6409.73 g/56 g/mol = 114.46 moles

Therefore, 114.46 moles of CO₂ were produces as well.

Molar volume of gas at STP = 22.4 litres

Volume of CO₂ produced at STP = 114.46 * 22.4 L =2563.904 L

However, the above reaction took place at 950 K and 0.976 atm, therefore volume of CO₂ produced under these conditions are obtained using the general gas equation

Using P₁V₁/T₁ = P₂V₂/T₂

P₁ = 1.0 atm, V₁ = 2563.904 L, T₁ = 273 K, P₂ = 0.976 atm, T₂ = 950 K, V₂ = ?

V₂ = P₁V₁T₂/P₂T₁

V₂ = (1.0 * 2563.904 * 950)/(0.976 * 273)

V₂ = 9141.404 L

Therefore, volume of CO₂ produced in the first step is 9141.404 L

3 0

3 years ago

Help me with these questions pls I will give brainliest

nirvana33 [79]

Question is irrelevant

7 0

2 years ago

What are some catchy volcano titles? I need it asap I'm making a 4th grade book

stich3 [128]

Volcano experiment
Volcano lavatory
Volcano extravaganza
Volcano blow thrash

3 0

4 years ago

Calculate the number of grams of sodium chloride in the solution. (Hint: Remember that sodium chloride is a strong electrolyte.)

creativ13 [48]

The question is incomplete, here is the complete question:

A solution contains 0.115 mol $H_2O$ and an unknown number of moles of sodium chloride. The vapor pressure of the solution at 30°C is 25.7 torr. The vapor pressure of pure water at this temperature is 31.8 torr. Calculate the number of grams of sodium chloride in the solution. (Hint: Remember that sodium chloride is a strong electrolyte.)

Answer: The mass of sodium chloride in the solution is 0.714 grams

Explanation:

The formula for relative lowering of vapor pressure will be:

$\frac{p^o-p_s}{p^o}=i\times \chi_{\text{solute}}$

where,

$p^o$ = vapor pressure of solvent (water) = 31.8 torr

$p^s$ = vapor pressure of the solution = 25.7 torr

i = Van't Hoff factor = 2

$\chi_{\text{solute}}$ = mole fraction of solute (sodium chloride) = ?

Putting values in above equation, we get:

$\frac{31.8-25.7}{31.8}=2\times \chi_{NaCl}\\\\\chi_{NaCl}=0.0959$

Mole fraction of a substance is calculated by using the equation:

$\chi_A=\frac{n_A}{n_A+n_B}$

$\chi_{\text{NaCl}}=\frac{n_{\text{NaCl}}}{n_{\text{NaCl}}+n_{\text{water}}}$

We are given:

Moles of water = 0.115 moles

$0.0959=\frac{n_{\text{NaCl}}}{n_{\text{NaCl}}+0.115}\\\\n_{\text{NaCl}}=0.0122mol$

To calculate the mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of NaCl = 0.0122 moles

Molar mass of NaCl = 58.5 g/mol

Putting values in above equation, we get:

$0.0122mol=\frac{\text{Mass of NaCl}}{58.5g/mol}\\\\\text{Mass of NaCl}(0.0122mol\times 58.5g/mol)=0.714g$

Hence, the mass of sodium chloride in the solution is 0.714 grams

5 0

3 years ago

How many grams of diphosphorus pentoxide result if 100.0 g of phosphorus are combined with sufficient oxygen?

gregori [183]

Answer is: 230 g.
Chemical reaction: P₄ + 5O₂ → 2P₂O₅.
m(P₄) = 100 g.
M(P₄) = 4 · 31 g/mol = 124 g/mol.
n(P₄) = m(P₄) ÷ M(P₄) = 100g ÷ 124g/mol = 0,806 mol.
From reaction: n(P₄) : n(P₂O5) = 1 : 2.
n(P₂O₅) = 1,612 mol.
m(P₂O₅) = 1,612 mol · 142g/mol = 230g.
M - molar mass.
n - amount of substance.

3 0

3 years ago