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harkovskaia [24]
3 years ago
6

Chandra wants to know how many moles are contained in 150.0g CCl4 what quantity of CCl4 should she use to make this conversion

Chemistry
2 answers:
DanielleElmas [232]3 years ago
7 0
Hey there!:

atomic mass :

C = 12.01070 amu

Cl = 35.4530 amu

Molar mass:

CCl4 = ( 1 * 12.01070 ) + ( 35.4530 *4 ) = 153.8227 g/mol

Therefore:

n = mass of solute / molar mass

n = 150.0 / 153.8227

n = 0.975 moles

hope this helps!
Orlov [11]3 years ago
5 0

Answer:

Molar mass.

Explanation:

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Answer:

Density of block of gold is <em>3.5 g/cm³.</em>

Explanation:

Given data:

Volume of block = 1000 mL

Mass of block = 3.5 kg (3.5×1000 = 3500 g)

Density of block = ?

Solution:

Density of substance is calculated by dividing the mass of substance over its volume.

Formula:

d = m/v

d = 3500 g/ 1000 mL

d = 3.5 g/mL

or

d = 3.5 g/cm³             (1ml = 1cm³)

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3 years ago
Equation: CH4 + 202 CO2 + 2H20
jeka94

Answer:

2:1

1.2 × 10² g

Explanation:

Step 1: Write the balanced combustion equation

CH₄ + 2 O₂ ⇒ CO₂ + 2 H₂O

Step 2: Establish the appropriate molar ratio

According to the balanced equation, the molar ratio of O₂ to CH₄ is 2:1.

Step 3: Calculate the moles of CH₄ required to react with 15 moles of O₂

We will use the previously established molar ratio.

15 mol O₂   1 mol CH₄/2 mol O₂ = 7.5 mol CH₄

Step 4: Calculate the mass corresponding to 7.5 moles of CH₄

The molar mass of CH₄ is 16.04 g/mol.

7.5 mol × 16.04 g/mol = 1.2 × 10² g

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The map shows watersheds with a high potential for soil, nitrogen, and pesticide runoff. If you were in charge of handling the p
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Answer:

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3 0
3 years ago
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If 500.0 mL of 0.10 M Ca2+ is mixed with 500.0 mL of 0.10 M SO42−, what mass of calcium sulfate will precipitate? Ksp for CaSO4
statuscvo [17]

Answer:

The mass of calcium sulfate that will precipitate is 6.14 grams

Explanation:

<u>Step 1:</u> Data given

500.0 mL of 0.10 M Ca^2+ is mixed with 500.0 mL of 0.10 M SO4^2−

Ksp for CaSO4 is 2.40*10^−5

<u>Step 2:</u> Calculate moles of Ca^2+

Moles of Ca^2+ = Molarity Ca^2+ * volume

Moles of Ca^2+ = 0.10 * 0.500 L

Moles Ca^2+ = 0.05 moles

<u>Step 3: </u>Calculate moles of SO4^2-

Moles of SO4^2- = 0.10 * 0.500 L

Moles SO4^2- = 0.05 moles

<u>Step 4: </u>Calculate total volume

500.0 mL + 500.0 mL = 1000 mL = 1L

<u>Step 5: </u> Calculate Q

Q = [Ca2+] [SO42-]  

[Ca2+]= 0.050 M   [O42-]

Qsp = (0.050)(0.050 )=0.0025 >> Ksp

This means precipitation will occur

<u> Step 6:</u> Calculate molar solubility

Ksp = 2.40 * 10^-5 = [Ca2+][SO42-] =(x)(x)

2.40 * 10^-5 = x²

x = √(2.40 * 10^-5)

x = 0.0049 M = Molar solubility

<u> Step 7:</u> Calculate total CaSO4 dissolved

total CaSO4 dissolved = 0.0049 M * 1 L * 136.14 mol/L = 0.667 g

<u>Step 8:</u> Calculate initial mass of CaSO4

Since initial moles CaSo4 = 0.050

Initial mass of CaSO4 = 0.050 * 136.14 g/mol

Initial mass of CaSO4 = 6.807 grams

<u>Step 9:</u> Calculate mass precipitate

6.807 - 0.667 = 6.14 grams

The mass of calcium sulfate that will precipitate is 6.14 grams

5 0
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