Determine the percent composition of phosphorous in calcium phosphate

A balloon contains 11.1 g of nitrogen monoxide gas (NO). how many molecules are in the balloon?

grandymaker [24]

The number of molecules that are in balloon are = 2.227 x10^23 molecules

<h3> calculation</h3>

calculate the number of moles of NO

moles = mass/molar mass

molar mass of NO = 14+ 16 = 30 g/mol

moles is therefore= 11.1 g/30g/mol= 0.37 moles

by use of Avogadro's constant that is

1 mole= 6.02 x10^23 molecules

0.37 =? molecules

=(6.02 x10^23 x 0.37 moles)/ 1mole=2.227 x10^23 molecules

4 0

3 years ago

Explain to me why a pressure cooker does not explode.

trapecia [35]

it can with stand certain pressure inside, once it exceeds the pressure will be released by making a whistle....... hence it wont burst

4 0

3 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]

<u>Answer:</u> The percentage by mass of NaCl in the mixture is 38.5 %

<u>Explanation:</u>

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)

<u>For NaCl:</u>

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

<u>For $NaNO_3$ :</u>

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 %

4 0

3 years ago

What is the pH of a 1 ´ 10-5 M KOH solution?

melisa1 [442]

KOH -------------> K⁺ + OH⁻

pOH = - log [ OH⁻ ]

pOH = - log [ 1 x 10⁻⁵ ]

pOH = 5

pH + pOH = 14

pH + 5 = 14

pH = 14 - 5

pH = 9

hope this helps!

3 0

3 years ago

A compound is found to contain 7.523% phosphorus and 92.48% iodine by weight. what is the empirical formula for this compound?

RUDIKE [14]

The answer is PI3
Ratio of 1:3

5 0

3 years ago

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