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Veseljchak [2.6K]
3 years ago
7

CF4 + Br2 CBr4 + F2

Chemistry
1 answer:
luda_lava [24]3 years ago
5 0

Answer:

Answers are in the explanation

Explanation:

Based on the reaction:

CF₄ + 2Br₂ → CBr₄ + 2F₂

The mole ratio of CF₄ is:

CF₄:Br₂ = 1:2

CF₄:CBr₄ = 1:1

CF₄:F₂ = 1:2

<em>Moles F2:</em>

Molar mass CF₄: 88.0g/mol

57.0g * (1mol / 88.0g) = 0.6477 moles CF₄ * (2mol F₂ / 1mol CBr₄) =

<h3>1.30 moles F₂</h3><h3 />

<em>Mass Br2:</em>

Molar mass CBr₄: 331.63g/mol

250.0g * (1mol / 331.63g) = 0.7539 moles CBr₄ * (2mol Br₂ / 1mol CF₄) =

1.51 moles Br₂ * (159.808g / mol) =

<h3>241g Br2</h3><h3 /><h3 />

<em>Moles F2:</em>

4.8 moles CF₄ * (2mol F₂ / 1mol CF₄) =

<h3>9.6 moles F₂</h3><h3 />

<em />

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Fossil A is found in a rock layer above a layer containing Fossil B. Which fossil is probably older?
ch4aika [34]

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4 0
3 years ago
How many molecules of O2 are in 153.9 g O2?
MrRissso [65]

Answer:

2.895*10^24

Explanation:

mass of Oxygen give = 153.9g

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n= mass/ molar mass

n= 153.9/32

n=4.81mol.

To find the number of molecules of o

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3 0
3 years ago
How many moles of water can be formed from 0.57 moles of hydrogen gas?
BartSMP [9]

Answer:

0.57 water

Explanation:

To solve this problem, we need to write the reaction expression first.

The reactants are oxygen gas and hydrogen gas.

They react to give a product of water

       2H₂    +    O₂   →   2 H₂O  

Given that;

Number of moles of hydrogen gas = 0.57moles

From the balanced reaction expression;

       2 moles of hydrogen gas produces 2 moles of water

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    0.57mole of hydrogen gas will also produce 0.57 water

7 0
2 years ago
Calculate the volume in ml of 0.200 M na2co3 needed to produce 2.00 g of caco3 there is an excess of cacl2
UkoKoshka [18]

Answer:

V=100mL

Explanation:

Hello.

In this case, since the chemical reaction is:

Na_2CO_3+CaCl_2\rightarrow  CaCO_3+2NaCl

We next compute the moles of sodium carbonate from the 2.00 grams of calcium carbonate via their 1:1 mole ratio in the chemical reaction:

n_{Na_2CO_3}=2.00gCaCO_3*\frac{1molCaCO_3}{100.09gCaCO_3}*\frac{1molNa_2CO_3}{1molCaCO_3}  \\\\n_{Na_2CO_3}=0.0200molNa_2CO_3

Thus, by knowing the molarity, we compute the volume:

M=\frac{n}{V}\\ \\V=\frac{n}{M}=\frac{0.0200mol}{0.200mol/L}\\  \\V=0.100L*\frac{1000mL}{1L}\\ \\V=100mL

Best regards.

8 0
3 years ago
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