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9966 [12]
3 years ago
15

0.39 moles of a gas fill a 5.0 l container that connects to a balloon through a closed valve. if the valve is opened an pressure

remains the same, what would be the total number of moles of gas needed in order to inflate the balloon by 1.5
Chemistry
1 answer:
harina [27]3 years ago
8 0

Answer: 5.39

Explanation: you are adding the balloon's volume and the container's volume

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What is the mass of 0.513 mol Al2O3? Give your answer to the correct number of significant figures. (Molar mass of Al2O3 = 102.0
prisoha [69]
.513mol x (102g/1mol)

Essentially, this is .513 x 102
Which equals: 52.326
But because you can only have 3 significant figures, your answer is:
52.3 grams

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7 0
3 years ago
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What is the pH of 0.000134 M solution of HCI?
mafiozo [28]

Answer: The pH will be 3.87

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

pH=-\log [H^+]

HCl\rightarrow H^++Cl^{-}

According to stoichiometry,

1 mole of HCl gives 1 mole of H^+

Thus 0.000134 moles of HCl gives =\frac{1}{1}\times 0.000134=0.0001342 moles of H^+

Putting in the values:

pH=-\log[0.000134]

pH=3.87

Thus the pH will be 3.87

5 0
3 years ago
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Al + AgNO3 -> Al(NO3)3 + Ag. As a balanced equation​
prisoha [69]

Answer:

Al + 4AgNO3 >>Al(NO3)3+ 3Ag

Explanation:

the number of moles of No3 of the products is 3 therefore we have to balance the reactants by adding 3 before the "AgNO3" which also leades us to adding 3 mols to Ag on the products side

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2 years ago
Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th
mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

where,

\Delta T_f = \text{Freezing point of pure solution}-\text{Freezing point of solution}

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

K_f = molal freezing point depression constant = 5.12 K/m  = 5.12 °C/m

m_{solute} = Given mass of solute (anthracene) = 7.99 g

M_{solute} = Molar mass of solute (anthracene) = 178.23  g/mol

W_{solvent} = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC

Hence, the freezing point of solution is 2.6°C

8 0
3 years ago
What is the mass of 4 × 1015 atoms of nickel<br> (Ni)?<br> Answer in units of g.
IRISSAK [1]

Answer:

4060

Explanation:

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