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Pani-rosa [81]
3 years ago
7

Determine the mass of 5.20 moles of C6H12

Chemistry
1 answer:
Vera_Pavlovna [14]3 years ago
6 0
Hello!
We have the following data:

m (mass) = ? 
n (number of moles) = 5.20 moles
MM (Molar mass of C6H12) 
C = 6*12 = 72 amu
H = 12*1 = 12 amu
--------------------------
MM (Molar mass of C6H12) = 72 + 12 = 84 g/mol

Now, let's find the mass, knowing that:

n =  \dfrac{m}{MM}

5.20\:\:\diagup\!\!\!\!\!\!\!mol =  \dfrac{m}{84\:g/\diagup\!\!\!\!\!\!\!mol}

m = 5.20*84

\boxed{\boxed{m = 436.8\:g}}\end{array}}\qquad\checkmark


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The stock name (CrBr3) :  <em>chromium(III) sulfide</em>

The classical name (CrBr3) : <em>chromic bromide</em>

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4 years ago
What is the numerical value of Kc for the following reaction if the equilibrium mixture contains 0.51 M
Goryan [66]

The  numerical value  of Kc  is 1.129  x10^4

 

<u><em>  Explanation</em></u>

C3H6O +4O2→  3 CO2 + 3H2O

KC  is the ratio  of  concentration of the   product  over  the reactant.

Each concentration of  product and  reactant   are raised  to the power  of  its coefficient.

Therefore the KC  expression  of  equation above  is

Kc=[ (Co2)^3 (H2O)^3]  / [(C3H6O) (O2)^4)]

 Kc  =[(1.8^3) x  (2.0^3)] / [(0.51)  x (0.30^4)] =1.129  x10^4


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How many moles of Cl– ions are in 2.20 L of 1.50 M of sodium-chloride aqueous solution?
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How many moles of Cl– ions are in 2.20 L of 1.50

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3 years ago
What is the volume at STP of 3.44 x 1023 molecules of CO2
almond37 [142]

Answer:

C. 12.8 liters.

Explanation:

The Standard Temperature and Pressure (STP) of a gas are 273.15 K and 100 kilopascals. From Avogadro's Law, a mole of carbon dioxide contains 6.022 \times 10^{23} molecules. If we suppose that carbon dioxide behaves ideally, then the equation of state for ideal gas is:

P\cdot V = n\cdot R_{u}\cdot T (1)

P\cdot V = \frac{r\cdot R_{u}\cdot T}{N_{A}} (1b)

Where:

P - Pressure, measured in pascals.

V - Volume, measured in liters.

r - Amount of molecules, no unit.

N_{A} - Avogadro's number, no unit.

R_{u} - Ideal gas constant, measured in pascal-liters per mole-Kelvin.

T - Temperature, measured in Kelvin.

If we know that P = 100000\,Pa, r = 3.44\times 10^{23}, N_{A} = 6.022\times 10^{23}, T = 273.15\,K and R_{u} = 8.314\times 10^{3}\,\frac{L\cdot Pa}{mol\cdot K}, then the volume of carbon dioxide at STP is:

V = \frac{r\cdot R_{u}\cdot T}{N_{A}\cdot P}

V = \frac{(3.44\times 10^{23})\cdot \left(8.314\times 10^{3}\,\frac{L\cdot Pa}{mol\cdot K} \right)\cdot (273.15\,K)}{(6.022\times 10^{23})\cdot (100000\,Pa)}

V = 12.972\,L

Therefore, the correct answer is C.

8 0
3 years ago
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