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3 years ago

Spectator ions when FeSO4 and PbNO32 react

Chemistry

1 answer:

sweet [91]3 years ago

4 0

Sorry I thought I had the answer!

Explanation:

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Got this really tricky extra credit question I can't figure out. Anyone want to give #24 a try?

agasfer [191]

Where is the pic at ???????/????

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3 years ago

What is the difference between heterochromatin and euchromatin.

qwelly [4]

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The main difference between the two is euchromatin is genetically active while heterochromatin is genetically inactive

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2 years ago

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An object has a mass of 10 grams and a volume of 5cm3 what is the density of the object?

lesya692 [45]

Answer:

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Explanation:

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2 years ago

Consider a mixture of NaCl and NaNO3 that is 31.8% Na by mass. Calculate the percentage by mass of NaCl in the mixture.

lutik1710 [3]

Answer: The percentage by mass of NaCl in the mixture is 38.5 %

Explanation:

To calculate the mass percentage of element in compound, we use the equation:

$\text{Mass percent of element}=\frac{\text{Mass of element}}{\text{Mass of compound}}\times 100$ .......(1)

For NaCl:

Mass of sodium element = 23 g

Mass of NaCl = 58.5 g

Putting values in equation 1, we get:

$\text{Mass percent of sodium element}=\frac{23g}{58.5g}\times 100=39.32\%$

Mass fraction of sodium metal in NaCl = 0.3932

For $NaNO_3$ :

Mass of sodium element = 23 g

Mass of $NaNO_3$ = 85 g

Putting values in equation 1, we get:

$\text{Mass percent of sodium element}=\frac{23g}{85g}\times 100=27.1\%$

Mass fraction of sodium metal in sodium nitrate = 0.271

Let us assume the mass fraction of NaCl in the mixture is 'x'

So, the mass fraction of $NaNO_3$ in the mixture will be '(1-x)'

We are given:

Percent by mass of Na in the mixture = 31.8 %

Mass fraction of Na in the mixture = 0.318

Evaluating the mass fraction of NaCl in the mixture:

$[(x\times 0.3932)+((1-x)\times 0.271)]=0.318$

x = 0.385

Percent by mass of NaCl in the mixture will be = $(0.385\times 100)=38.5\%$

Hence, the percentage by mass of NaCl in the mixture is 38.5 %

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3 years ago

How many joules of heat are absorbed to raise the temperature of 435 grams of water at 1 atm from 25°C to its boiling point, 100

rosijanka [135]

136568.25 J

Let me know if you have any questions

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4 years ago