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3 years ago

What is the approximate volume of 19g of fluorine gas at a pressure of 4.0 atmospheres and a temperature of 127o C?

Chemistry

1 answer:

mr Goodwill [35]3 years ago

5 0

Answer:

V = 8.21 L

Explanation:

Given data:

Volume of fluorine = ?

Mass of fluorine = 19 g

Pressure = 4.0 atm

Temperature = 127 °C(127+ 273 =400 K)

Solution:

Number of moles of fluorine:

Number of moles = mass/molar mass

Number of moles = 19 g/ 19 g/mol

Number of moles = 1 mol

PV = nRT

V = nRT/P

V = 1 mol ×0.0821 atm. L. mol⁻¹. k⁻¹ × 400 K/ 4 atm

V = 32.84 L/ 4

V = 8.21 L

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which of the following is a physical change? 1.burning paper 2.grinding wheat 3.electrolysis of water 4.cooking rice

azamat

Answer:

2 grinding of wheat is just a physical change

plz brainlist

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2 years ago

Can Iron replace Hydrogen in the following reaction?: Fe(s) + HCl(aq) ---> H2(g) + FeCl2(aq)

Elenna [48]

It can because Iron is more reactive than Hydrogen and a more reactive metal always replaces a less reactive one in a single replacement reaction

3 0

3 years ago

Calculate the standard entropy of vaporization of ethanol at its boiling point, 352 K. The standard molar enthalpy of vaporizati

Korvikt [17]

Answer : The correct option is, (b) +115 J/mol.K

Explanation :

Formula used :

$\Delta S=\frac{\Delta H_{vap}}{T_b}$

where,

$\Delta S$ = change in entropy

$\Delta H_{vap}$ = change in enthalpy of vaporization = 40.5 kJ/mol

$T_b$ = boiling point temperature = 352 K

Now put all the given values in the above formula, we get:

$\Delta S=\frac{\Delta H_{vap}}{T_b}$

$\Delta S=\frac{40.5kJ/mol}{352K}$

$\Delta S=115J/mol.K$

Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K

8 0

3 years ago

Does anyone know how to do this? the solubility of barium carbonate, BaCO3, is 0.0100 g/L. Its molar mass is 197.3 g/mol. What i

g100num [7]

It's simple, just follow my steps.

1º - in 1 L we have $0.0100~g$ of $BaCO_3$

2º - let's find the number of moles.

$\eta=\frac{m}{MM}$

$\eta=\frac{0.0100}{197.3}$

$\boxed{\boxed{\eta=5.07\times10^{-5}~mol}}$

3º - The concentration will be

$C=5.07\times10^{-5}~mol/L$

But we have this reaction

$BaCO_3\rightleftharpoons Ba^{2+}+CO_3^{2-}$

This concentration will be the concentration of $Ba^{2+}~~and~~CO_3^{2-}$

$K_{sp}=\frac{[Ba^{2+}][CO_3^{2-}]}{[BaCO_3]}$

considering $[BaCO_3]=1~mol/L$

$K_{sp}=[Ba^{2+}][CO_3^{2-}]$

and

$[Ba^{2+}]=[CO_3^{2-}]=5.07\times10^{-5}~mol/L$

We can replace it

$K_{sp}=(5.07\times10^{-5})*(5.07\times10^{-5})$

$K_{sp}\approx25.70\times10^{-10}$

Therefore the $K_{sp}$ is:

$\boxed{\boxed{\boxed{K_{sp}\approx2.57\times10^{-11}}}}$

4 0

3 years ago

Read 2 more answers

If a penny (with the this volume .39 cm^3) was made out of pure copper, what should its mass be?

blondinia [14]

Answer:

3.49 g

Explanation:

The mass is the product of volume and density:

(8.96 g/cm³)(0.39 cm³) ≈ 3.49 g

The mass of a pure-copper penny would be 3.49 g.

5 0

2 years ago