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topjm [15]
2 years ago
6

How many moles are in 11.5 L of a gas at STP?

Chemistry
1 answer:
garik1379 [7]2 years ago
6 0

Answer:

0.51mole

Explanation:

Given parameters:

Volume of gas  = 11.5L

Unknown:

Number of moles = ?

Solution:

Since the gas is at STP,

     Number of moles  = \frac{volume of the gas }{22.4}  

    Insert parameters:

 Number of moles  = \frac{11.5}{22.4}   = 0.51mole

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the goal of any scientific method is to solve a problem or to better understand an observed event.

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i. What does the law of conservation of energy say?

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The law of conservation of energy states that energy can neither be created nor destroyed - only converted from one form of energy to another.

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Potential energy.

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3 years ago
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Juanita dissolves 46 g of MgBr2 (molar mass: 184.11 g/mol) in 0.5 kg of distilled water. What is the molality of the solution?
irina [24]
Formula: molality, m = n solute / kg solvent

n solute = # of moles of solute = mass(g) / molar mass

Molar mass of Mg Br2 = 184.11 g/mol

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3 years ago
An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for
8090 [49]

Answer: The freezing point and boiling point of the solution are -6.6^0C and 101.8^0C respectively.

Explanation:

Depression in freezing point:

T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_f = freezing point of solution = ?

T^o_f = freezing point of water = 0^0C

k_f = freezing point constant of water = 1.86^0C/m

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

w_2 = mass of solute (ethylene glycol) = 21.4 g

w_1= mass of solvent (water) = density\times volume=1.00g/ml\times 97.6ml=97.6g

M_2 = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}

T_f=-6.6^0C

Therefore,the freezing point of the solution is -6.6^0C

Elevation in boiling point :

T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}

where,

T_b = boiling point of solution = ?

T^o_b = boiling point of water = 100^0C

k_b = boiling point constant of water = 0.52^0C/m

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

w_2 = mass of solute (ethylene glycol) = 21.4 g

w_1= mass of solvent (water) = density\times volume=1.00g/ml\times 97.6ml=97.6g

M_2 = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}

T_b=101.8^0C

Thus the boiling point of the solution is 101.8^0C

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Anvisha [2.4K]

Answer:

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Answer:

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