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3 years ago

What is the molarity of 2.34 mil of calcium in 1860 mL of solution?

Chemistry

2 answers:

jenyasd209 [6]3 years ago

7 0

Answer: 1.23 M Ca

Explanation:

Molarity is moles of solute/volume (only liters) of solution

Convert mL to L

To do that divide by 1000

Therefore, 2.34 mol/1.86 L

Answer is 1.23 M Ca

shtirl [24]3 years ago

4 0

Answer:

1.258 M

Explanation:

Molarity = $\frac{Moles of solute}\\{Volume of the solution in L}$

Given.

Moles of Calcium = 2.34 mol

Volume of the solution = 1860 ml

1 L = 1000 ml

1860 ml x $\frac{1L}{1000 ml}$ = 1.86 L

Molarity = $\frac{2.34 mol}{1.86 L}$ = 1.258 M

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Consider the followong balanced reaction. What mads in g of co2 can be formed from 288 mg of o2. Assume that there is excess c3h

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Answer:

0.264 g of $CO_2$ can be formed from 288 mg of $O_2$

Explanation:

The balanced chemical equation is

$2 C_3 H_7 OH+9 O_2> 6 CO_2+8 H_2 O$

The conversions are

Mass in mg $O_2$ is converted to mass in g $O_2$

Mass in g $O_2$ is converted to moles $O_2$ by dividing with molar mass

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Moles $CO_2$ is converted to mass $CO_2$ by multiplying with molar mass $CO_2$

mass in mg $O_2$ > mass in g $O_2$ >moles $O_2$ > moles $CO_2$ > mass $CO_2$

$288mg O_2 \times \frac{(1g O_2)}{(1000mg O_2 )} \times \frac {(1molO_2)}{(32gO_2 )}\times\frac {(6mol CO_2)}{(9mol O_2 )} \times \frac {(44.0 gCO_2)}{(1mol CO_2 )}$

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3 years ago

Sodium metal reacts with water to produce hydrogen gas and sodium hydroxide according to the chemical equation shown below.

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Answer: The enthalpy change of the reaction is -361.6 kJ

Explanation:

To calculate the number of moles, we use the equation:

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

Moles of sodium = 0.025 moles

Molar mass of sodium = 23 g/mol

Putting values in above equation, we get:

$0.025mol=\frac{\text{Mass of sodium}}{23g/mol}\\\\\text{Mass of sodium}=(0.025mol\times 23g/mol)=0.575g$

We are given:

Mass of water = 100.00 g

Mass of sodium = 0.575 g

Mass of solution = 100.00 + 0.575 = 100.575 g

To calculate the amount of heat absorbed, we use the equation:

$q=m\times C\times \Delta T$

where,

q = amount of heat absorbed = ?

m = mass of solution = 100.575 g

C = specific heat capacity of solution = 4.18 J/g°C

$\Delta T$ = change in temperature = $(T_2-T_1)=(35.75-25.00)=10.75^oC$

Putting all the values in above equation, we get:

$q=100.575g\times 4.18J/g^oC\times 10.75^oC=4519.34J=4.52kJ$

When heat is absorbed by the solution, this means that heat is getting released by the reaction.

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

For the given chemical reaction:

$2Na(s)+2H_2O(l)\rightarrow NaOH(aq.)+H_2(g)$

When 0.025 moles of sodium is reacted, the heat released by the reaction is 4.52 kJ

So, when 2 moles of sodium will react, the heat released by the reaction will be = $\frac{4.52}{0.025}\times 2=361.6kJ$

Hence, the enthalpy change of the reaction is -361.6 kJ

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