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2 years ago

5

A graduated cylinder initially contained 11 mL of water and then an 8 g block of wood is put into it, changing the volume to 15

mL. What is the density of the wood?

1 answer:

Paul [167]2 years ago

8 0

Answer:

Density is 2

Explanation:

Do the final volume of water divided by the inital

volume of water to get 4

15-11 = 4

Volume = 2

Mass = 8

Then to the equation m/v

8/4 = 2

Density = 2

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Which type of stoichiometry problems does not require the use of molar mass?

Arlecino [84]

Answer:

Explanation:

It is volume-volume problems that does not require the use of molar mass.

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2 years ago

A certain substance X has a normal freezing point of -6.4 C and a molal freezing point depression constant Kf= 3.96 degrees C.kg

Brut [27]

Answer: $1.0\times 10^2g$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(-6.4-(13.6))^0C=7.2^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte like urea)

$K_f$ = freezing point constant = $3.96^0C/m$

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}}\times \text{weight of solvent in kg}}$

Weight of solvent (X)= 950 g = 0.95 kg

Molar mass of non electrolyte (urea) = 60.06 g/mol

Mass of non electrolyte (urea) added = ?

$7.2=1\times 3.96\times \frac{xg}{60.06 g/mol\times 0.95kg}$

$x=1.0\times 10^2g$

Thus $1.0\times 10^2g$ urea was dissolved.

8 0

3 years ago

A sample of ideal gas is in a sealed container. The pressure of the gas is 485 torr , and the temperature is 40 ∘C . If the temp

Fantom [35]

Answer:

537.68 torr.

Explanation:

We can use the general law of ideal gas:<em> PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and V are constant, and have different values of P and T:

<em>(P₁T₂) = (P₂T₁).</em>

P₁ = 485 torr, T₁ = 40°C + 273 = 313 K,

P₂ = ??? torr, T₂ = 74°C + 273 = 347 K.

∴ P₂ = (P₁T₂)/(P₁) = (485 torr)(347 K)/(313 K) = 537.68 torr.

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2 years ago

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scoray [572]

A. heat; light energies

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Co2 is correct buddy

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