Chemistry, 29.01.2020 14:43 haileycrane

What mass of naf must be mixed with 25ml of water to create a 3.5% by mass solution

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Answer from: thereal6ix9ine69

0.91 g NaF

Explanation:

Mass percentage is one way of representing the concentration of an element in a compound or a component in a mixture. Mass percentage is calculated as the mass of a component divided by the total mass of the mixture, multiplied by 100%.

Let mass of NaF = x g

∴ mass percent = mass of NaF / mass of total mixture *100

3.5 = (x/ x + 25) * 100

∴ x = 0.91 g

Answer from: n987

The mass of NaF that must be mixed with 25 mL of water is 0.91 grams

Explanation:

We are given:

Volume of water = 25 mL

To calculate the mass of water, we use equation used to calculate density:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 25 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{25mL}\\\\\text{Mass of water}=(1g/mL\times 25mL)=25g$

We are given:

3.5 % (m/m) NaF solution

This means that 3.5 grams of NaF is present in 100 g of solution

Mass of water in solution = 100 - 3.5 = 96.5 g

To calculate the mass of NaF mixed with 25 mL of water, we use unitary method:

In 96.5 grams of water, the mass of NaF added is 3.5 grams

So, in 25 grams of water, the mass of NaF that must be added will be = $\frac{3.5}{96.5}\times 25=0.91g$

Hence, the mass of NaF that must be mixed with 25 mL of water is 0.91 grams

Answer from: Quest

Change in speed or direction : )

Answer from: Quest

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 : $4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

equation 2 : $2x (s) + 3cl_{2} (g) \rightarrow 2xcl_{3} (s) h rxn = -800 kj/mol$

equation 3 : $4xcl_{3} (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) + 6cl_(2} h rxn = -200 kj/mol$

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 : $4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+800) \times 2 kj/mol$

let us add equation 1 & 4

$4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

$4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+1600) kj/mol$

$4xcl_{3} )s) + 3o_{2} \rightarrow 2x_{2}o_{3} (s) + 6cl_{2} (g) h rxn = (+1000) kj/mol$

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

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What mass of naf must be mixed with 25ml of water to create a 3.5% by mass solution...

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