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2 years ago

What volume in litres will 38gr of F2 occupy at 0.999 bar and 273 K

Chemistry

1 answer:

FrozenT [24]2 years ago

7 0

Answer:

$V=22.68L$

Explanation:

Hello,

In this case, we use the ideal gas equation to compute the volume as shown below:

$PV=nRT\\\\V=\frac{nRT}{P}$

Nonetheless we are given mass, for that reason we must compute the moles of gaseous fluorine (molar mass: 38 g/mol) as shown below:

$n=38 g *\frac{x}{y} \frac{1mol}{38 g} =1mol$

Thus, we compute the volume with the proper ideal gas constant, R:

$V=\frac{1mol*0.083\frac{bar*L}{mol*K}*273K}{0.999bar} \\\\V=22.68L$

Best regards.

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Summary about helium

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Answer: Helium is an type of air that can be used to inflate balloons, balls, and other things that require air. The carbon dioxide we breath out is similar to this helium.

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2 years ago

Look at the diagram below. What type of bond does it show?

MrRissso [65]

Explanation:

it shows covalent bonds

6 0

2 years ago

Determine the theoretical yield and the percent yield if 21.8 g of K2CO3 is produced from reacting 27.9 g KO2 with 57.0 g CO2. T

Alex_Xolod [135]

Answer:

26.9 g

81%

Explanation:

The equation of the reaction is;

4 KO2(s) + 2 CO2(g) → 3 O2(g) + 2 K2CO3(s)

Number of moles of KO2= 27.9g/71.1 g/mol = 0.39 moles

4 moles of KO2 yields 2 moles of K2CO3

0.39 moles of KO2 yields 0.39 × 2/4 = 0.195 moles of K2CO3

Number of moles of CO2 = 57g/ 44.01 g/mol = 1.295 moles

2 moles of CO2 yields 2 moles of K2CO3

1.295 moles of CO2 yields 1.295 × 2/2 = 1.295 moles of K2CO3

Hence the limiting reactant is KO2

Theoretical yield = 0.195 moles of K2CO3 × 138.205 g/mol = 26.9 g

Percent yield = actual yield/theoretical yield × 100

Percent yield = 21.8/26.9 × 100

Percent yield = 81%

5 0

3 years ago

How many moles of oxygen (O) atoms are in a sample of 3.98 × 1023 atoms? 0.661 mol O 1.51 mol O 6.60 × 1022 mol O 2.40 × 1047 mo

lidiya [134]

To find moles from any number of atoms, always divide by the constant called Avogadro's Number. This number is representative of the number of atoms or molecules in a mole of any substance and is equal to about $6.022* 10^{23} 
$ .

So, for this problem, we take the number of atoms of oxygen and divide by Avogadro's Number.

$\frac{3.98* 10^{23} O_{2} atoms }{6.022* 10^{23} atoms/mol } = 0.6609 moles$

Answer: 0.661 moles (after rounding)

Make sense?

7 0

3 years ago

Solid calcium carbide, CaC2, reacts with water to form an aqueous solution of calcium hydroxide and acetylene gas, C2H2. Express

Lina20 [59]

Answer: $CaC_2(s)+H_2O(l)\rightarrow Ca(OH)_2(aq)+C_2H_2(g)$

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The solutions in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas. Liquids are represented by (l) and gases are represented by (g).

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3 years ago