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3 years ago

10

If Earth were not tilted on its axis, what change would we experience? A) Day length would be less than 24 hours. B) Both sides

of the moon would be visible. C) Temperatures would be constant year round. D) The length of the year would be more than 365 days.

2 answers:

77julia77 [94]3 years ago

4 0

Temperatures would be constant year round

Bezzdna [24]3 years ago

4 0

The answer is C

God bless

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During dry weather, the water level in a small pond drops because of

Sophie [7]

Water levels tend to drop in dry weather because of evaporation. The hot temperatures will cause the water to achieve its gaseous state and go up into the atmosphere.

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2 years ago

Consider the reaction between iron (III) oxide, Fe2O3 and carbon monoxide, CO.

Lana71 [14]

<span>Fe2O3 + 3CO --> 2Fe + 3CO2
</span><span>
m(Fe2O3)=213 g
m(CO)=140 g
</span>_______________

<span>n(Fe2O3)=?
m(Fe)=?
n(Fe2O3)=?
n(CO)=?
n(CO2)=?
</span>
<span>n(Fe2O3)=m(Fe2O3) / M(Fe2O3)
n(Fe2O3)= 213 g / 159,7 gmol-1 = 1,33 mol
</span>
<span>n(CO)= m(CO) / M(CO) n(CO)= 140 g / 28,01 gmol-1 = 4,99 mol</span>

3 0

3 years ago

When of alanine are dissolved in of a certain mystery liquid , the freezing point of the solution is less than the freezing poin

LenaWriter [7]

The question is incomplete, the complete question is:

When 177. g of alanine $(C_3H_7NO_2)$ are dissolved in 800.0 g of a certain mystery liquid X, the freezing point of the solution is $5.9^oC$ lower than the freezing point of pure X. On the other hand, when 177.0 g of potassium bromide are dissolved in the same mass of X, the freezing point of the solution is $7.2^oC$ lower than the freezing point of pure X. Calculate the van't Hoff factor for potassium bromide in X.

<u>Answer:</u> The van't Hoff factor for potassium bromide in X is 1.63

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

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

OR

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

<u>When alanine is dissolved in mystery liquid X:</u>

$\Delta T_f=5.9^oC$

i = Vant Hoff factor = 1 (for non-electrolytes)

$K_f$ = freezing point depression constant

$m_{solute}$ = Given mass of solute (alanine) = 177. g

$M_{solute}$ = Molar mass of solute (alanine) = 89 g/mol

$w_{solvent}$ = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

$5.9=1\times K_f\times \frac{177\times 1000}{89\times 800}\\\\K_f=\frac{5.9\times 89\times 800}{1\times 177\times 1000}\\\\K_f=2.37^oC/m$

<u>When KBr is dissolved in mystery liquid X:</u>

$\Delta T_f=7.2^oC$

i = Vant Hoff factor = ?

$K_f$ = freezing point depression constant = $2.37^oC/m$

$m_{solute}$ = Given mass of solute (KBr) = 177. g

$M_{solute}$ = Molar mass of solute (KBr) = 119 g/mol

$w_{solvent}$ = Mass of solvent = 800.0 g

Putting values in equation 1, we get:

$7.2=i\times 2.37\times \frac{177\times 1000}{119\times 800}\\\\i=\frac{7.2\times 119\times 800}{2.37\times 177\times 1000}\\\\i=1.63$

Hence, the van't Hoff factor for potassium bromide in X is 1.63

7 0

3 years ago

What is the volume of a gas if 0.182 moles of the gas is at 1.99 atm and 83.4oC?

Pie

$\text{Given that,}\\\\\text{Pressure,}~ P = 1.99~ \text{atm}\\ \\\text{No. of moles,}~ n = 0.182~ \text{mol.}\\\\\text{Temperature,}~ T = 83.4^{\circ}~ C = 356.4~K\\\\ \text{Molar gas constant,}~ R = 0.082~ \text{L}~ \text{atm}~ \text{mol}^{-1}~ \text{K}^{-1}\\\\\text{Volume,}~ V =?\\\\\text{We know that,}~\\\\~~~~~~~PV =nRT\\\\\implies V = \dfrac{nRT}{P}\\\\\\~~~~~~~~~~=\dfrac{0.182 \times0.082 \times 356.4 }{1.99}\\\\\\~~~~~~~~~~\approx 2.68~ \text{L.}\\\\\\$

$\text{Hence the volume of the gas is 2.68 L.}$

6 0

1 year ago

Read 2 more answers

Calculate the ΔG°' for the reaction with 3 significant figures with no label for the dimension (just the number). fructose-6-pho

seraphim [82]

Answer:

ΔG° = 1747.523

Explanation:

The parameters mentioned are;

Gibbs Free energy ΔG°

Equilibrium constant Kc

Temperature T = 37 + 273 = 310 (upon conversion to kelvin temperature)

The formular relating all three parameters is given as;

ΔG° = -RTlnKc

Where; R = rate constant = 8.314 J⋅K−1⋅mol−1

Upon solving;

ΔG° = - 8.314 * 310 * ln(1.97)

ΔG° = 1747.523

6 0

3 years ago