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kipiarov [429]
2 years ago
9

Relative atomic mass of tin sulphide​

Chemistry
2 answers:
sweet-ann [11.9K]2 years ago
8 0

Answer:

150

Explanation:

Serga [27]2 years ago
6 0
Theoretical)
Compound Formula S2Sn
Density 4.5 g/cm3
Solubility in H2O Insoluble
Crystal Phase / Structure Rhombohedral
Exact Mass 183.846337
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A 13.5 g sample of an unknown gas occupies 5.10 L at 149.83 kPa and 301 K. What is the molar mass of the gas ?
alisha [4.7K]

Answer:

The molar mass of the gas is 44.19 g/mol

Explanation:

Amount of sample of gas = m = 13.5 g

Volume occupied by the gas = V = 5.10 L

Pressure of the gas = P = 149.83 KPa

1 KPa = 0.00986 atm

P = 149.83 \textrm{ KPa} \times 0.00986 \textrm{ atm/KPa} = 1.48 \textrm{ atm}

Assuming M g/mol to be the molar mass of the gas

Assuming the gas is behaving as an ideal gas

\textrm{PV} =\textrm{nRT} \\\textrm{PV} = \displaystyle \frac{m}{M}\textrm{ RT } \\1.48 \textrm{ atm}\times 5.10 \textrm{ L} = \displaystyle \frac{13.5 \textrm{ g}}{M}\times 0.0821 \textrm{ L.atm.mol}^{-1}.K^{-1}\times 301\textrm{K} \\M = 44.19 \textrm{ g/mol}

The molar mass of gas is 44.19 g/mol

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3 years ago
Explain in terms of particles in the atoms of an element why statement c is false?
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Explain the general process of nuclear fission. What is created from fission?
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In order for a solute to dissolve in a solvent, what must be true?
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2 years ago
Will GIVE BRAINLIEST --A student makes a standard solution of potassium hydroxide by adding 14.555 g to 500.0 mL of water. Answe
leva [86]

Answer:

0.5188 M or 0.5188 mol/L

Explanation:

Concentration is calculated as <u>molarity</u>, which is the number of moles per litre.

***Molarity is represented by either "M" or "c" depending on your teacher. I will use "c".

The formula for molarity is:

c = \frac{n}{V}

n = moles (unit mol)

V = volume (unit L)

<u>Find the molar mass (M) of potassium hydroxide.</u>

M_{KOH} = \frac{39.098 g}{mol}+\frac{16.000 g}{mol}+\frac{1.008 g}{mol}

M_{KOH} = 56.106 \frac{g}{mol}

<u>Calculate the moles of potassium hydroxide.</u>

n_{KOH} = \frac{14.555 g}{1}*\frac{1mol}{56.106g}

n_{KOH} = 0.25941(9)mol

Carry one insignificant figure (shown in brackets).

<u>Convert the volume of water to litres.</u>

V = \frac{500.0mL}{1}*\frac{1L}{1000mL}

V = 0.5000L

Here, carrying an insignificant figure doesn't change the value.

<u>Calculate the concentration.</u>

c = \frac{n}{V}

c = \frac{0.25941(9)mol}{0.5000 L}              

c = 0.5188(3) \frac{mol}{L}         <= Keep an insignificant figure for rounding

c = 0.5188 \frac{mol}{L}              <= Rounded up

c = 0.5188M               <= You use the unit "M" instead of "mol/L"

The concentration of this standard solution is 0.5188 M.

7 0
3 years ago
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