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3 years ago

How many grams of N are in 89.0 g of NH4NO3

1 answer:

8 0

Step 1;calculate the molar mass of NH4N03
molar mass of different element
N=14
H=1
O=16
therefore molar mass of NH4NO3 is 14+4+14+48= 80g/mol
step 2:find the molar mass of N present in NH4NO3
That is 14+14 =28g/mol
the mass is therefore
{(28g/mol/80g/mol) x 89g} =31.15g

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How many molecules are contaiined in 125 grams of water, H2O?

vladimir1956 [14]

Answer:

Water has a molar mass of 18.015 g/mol . This means that one mole of water molecules has a mass of 18.015 g . So, to sum this up, 6.022⋅1023 molecules of water will amount to 1 mole of water, which in turn will have a mass of 18.015 g . 2.7144moles H2O ⋅6.022⋅1023molec.

Explanation:

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3 years ago

CaBr + KOH – Ca(OH), + KBr (balance first) What mass, in grams, of

neonofarm [45]

Answer:

129.73 g of CaBr₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

CaBr₂ + 2KOH –> Ca(OH)₂ + 2KBr

Next, we shall determine the mass of CaBr₂ that reacted and the mass of Ca(OH)₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of CaBr₂ = 40 + (80×2)

= 40 + 160

= 200 g/mol

Mass of CaBr₂ from the balanced equation = 1 × 200 = 200 g

Molar mass of Ca(OH)₂ = 40 + 2(16 + 1)

= 40 + 2(17)

= 40 + 34

= 74 g/mol

Mass of Ca(OH)₂ from the balanced equation = 1 × 74 = 74 g

SUMMARY :

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Finally, we shall determine the mass of CaBr₂ that react when 48 g of Ca(OH)₂ were produced. This can be obtained as follow:

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Therefore, Xg of CaBr₂ will react to produce 48 g of Ca(OH)₂ i.e

Xg of CaBr₂ = (200 × 48)/74

Xg of CaBr₂ = 129.73 g

Thus, 129.73 g of CaBr₂ were consumed.

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3 years ago

What is the solution to the problem expressed to the correct number of significant figures

quester [9]

I don't know what the problem is, but here are some rues to help you out:

All non-zero figures are significant
When a zero falls between non-zero digits, that zero is significant.
When a zero falls after a decimal point, that zero is significant.
When multiplying and dividing significant figures, the answer is limited to the number of sig figs equal to the least number of sig figs in the problem.
When adding and subtracting, the answer is limited to the number of decimal places in the number with the least number of decimal places.

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3 years ago

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nikitadnepr [17]

Decantation is a process for the separation of mixtures of immiscible liquids or of a liquid and a solid mixture such as a suspension.

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3 years ago

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OLga [1]

Answer:

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Explanation:

can i have a brainlist :)

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3 years ago

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