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Tcecarenko [31]
3 years ago
13

Balance this reaction:C6H10Cl4+Cl2=​

Chemistry
1 answer:
Valentin [98]3 years ago
5 0

Answer:

C6H10 + 2Cl2 → C6H10Cl4

Explanation:

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At 25°C, gas in a rigid cylinder with a movable piston has a volume of 145 mL and a pressure of 125 kPa. Then the gas is compres
natta225 [31]

Answer:

(4) 230 kPa

Explanation:

The temperature is constant, so the only variables are pressure and volume.

We can use Boyle’s Law.

p_{1}V_{1} = p_{2}V_{2}

Divide both sides of the equation by V_{2}.

p_{2} = p_{1} \times \frac{V_{1}}{ V_{2}}

p_{1} = \text{145 kPa}; V_{1} = \text{125 cm}^{3}}

p_{2} = ?; V_{2} = \text{80. cm}^{3}

p_{2} = \text{145 kPa} \times \frac{\text{125 cm}^{3}}{\text{80. cm}^{3}} = \textbf{230 kPa}

7 0
3 years ago
How many microliters are in 7.4 x 10^-61 centimeters?
schepotkina [342]

Answer:

Microliters = 7.4\times 10^{-58}

Explanation:

SI unit of volume is liter.

1 dm^{3}=1 liter ..............................(1)

1dm=10cm

1dm^{3} = 1000 cm^{3}.................(2)

replacing value of 1 dm^{3}from equation (1) in equation (2)

1 liter = 1000 cm^{3}...................(3)

but

1 liter = 10^{6} microliters....................(4)

replacing value of 1 liter from equation (3) in equation (4)

1000 cm^{3} = 10^{6} microliters

on solving further ,we get

1 cm^{3} = \frac{10^{6}}{10^{3}}microliter

1 cm^{3} = 1000 microliter

7.4\times 10^{-61}\times 1000

Microliters = 7.4\times 10^{-58}

8 0
3 years ago
Pls help asap!!! BrainIdentify one source of carbon that can be found in each of these four spheres: atmosphere (air), geosphere
dedylja [7]

Answer:

It is carbon dioxide

Explanation:

<h2>All source relies on it Because if they dont have it then its not considered a source of anything else.</h2>
4 0
3 years ago
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What bond forms from an unequal sharing of electrons between atoms in a molecule? a nonpolar covalent bond a hydrophobic interac
love history [14]
Sharing of electrons always means its a covalent bond, and unequal means it is polar, so it is a polar covalent bond
7 0
4 years ago
Consider the combustion of octane (C8H18)
sesenic [268]
The balanced equation for the above reaction is as follows;
2C₈H₁₈ + 25O₂ --->  16CO₂  + 18H₂O
stoichiometry of octane to CO₂ is 2:16
number of C₈H₁₈ moles reacted - 191.6 g / 114 g/mol = 1.68 mol
when 2 mol of octane reacts it forms 16 mol of CO₂
therefore when 1.68 mol of octane reacts - it forms 16/2 x 1.68 = 13.45 mol of CO₂
number of CO₂ moles formed - 13.45 mol
therefore mass of CO₂ formed - 13.45 mol x 44 g/mol = 591.8 g
 mass of CO₂ formed is 591.8 g
3 0
3 years ago
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