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kobusy [5.1K]
3 years ago
8

Determine the number of moles of C in each sample

Chemistry
1 answer:
stepan [7]3 years ago
7 0

<u>Explanation:</u>

  • <u>In </u>CH_4<u>:</u>

Given moles = 2.0 moles

1 mole of methane contains 1 mole of carbon and 4 moles of hydrogen

Moles of carbon in CH_4=(2.0\times 1)=2.0 moles

  • <u>In </u>C_2H_6<u>:</u>

Given moles = 0.175 moles

1 mole of ethane contains 2 moles of carbon and 6 moles of hydrogen

Moles of carbon in C_2H_6=(0.175\times 2)=0.35 moles

  • <u>In </u>C_4H_{10}<u>:</u>

Given moles = 4.21 moles

1 mole of butane contains 4 moles of carbon and 10 moles of hydrogen

Moles of carbon in C_4H_{10}=(4.21\times 4)=16.84 moles

  • <u>In </u>C_8H_{18}<u>:</u>

Given moles = 24.5 moles

1 mole of octane contains 8 moles of carbon and 18 moles of hydrogen

Moles of carbon in C_8H_{18}=(24.5\times 8)=196 moles

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Kay [80]
You just need to multiply the total mass by the decimal value of the part that is tin. 133.8*0.103=13.8g (following the rules of significant figures).
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4 years ago
the speed limit is posted as 35 km/hr . your speedometer reads that you are going 40 miles/hr. are you speeding?
Morgarella [4.7K]
No, because 40 miles is the same as nearly 25 km/h. 
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3 years ago
Can somebody help me, please!
zavuch27 [327]

Answer:

Rubidium-85=61.2

Rubidium-87=24.36

Atomic Mass=85.56 amu

Explanation:

To find the atomic mass, we must multiply the masses of the isotope by the percent abundance, then add.

<u>Rubidium-85 </u>

This isotope has an abundance of 72%.

Convert 72% to a decimal. Divide by 100 or move the decimal two places to the left.

  • 72/100= 0.72      or        72.0 --> 7.2 ---> 0.72

Multiply the mass of the isotope, which is 85, by the abundance as a decimal.

  • mass * decimal abundance= 85* 0.72= 61.2

Rubidium-85=61.2

<u>Rubidium-87</u>

This isotope has an abundance of 28%.

Convert 28% to a decimal. Divide by 100 or move the decimal two places to the left.

  • 28/100= 0.28       or        28.0 --> 2.8 ---> 0.28

Multiply the mass of the isotope, which is 87, by the abundance as a decimal.

  • mass * decimal abundance= 87* 0.28= 24.36

Rubidium-87=24.36

<u>Atomic Mass of Rubidium:</u>

Add the two numbers together.

  • Rb-85 (61.2) and Rb-87 (24.36)
  • 61.2+24.36=85.56 amu
4 0
3 years ago
For the reaction Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3 how many grams of calcium carbonate, CaCO3, are produced from 79.3 g of sodium car
Alexus [3.1K]

Answer:

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

Explanation:

The balanced reaction is:

Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • Na₂CO₃: 1 mole
  • Ca(NO₃)₂: 1 mole
  • CaCO₃: 1 mole
  • NaNO₃: 2 mole

Being the molar mass of the compounds:

  • Na₂CO₃: 106 g/mole
  • Ca(NO₃)₂: 164 g/mole  
  • CaCO₃: 100 g/mole
  • NaNO₃: 85 g/mole

then by stoichiometry the following quantities of mass participate in the reaction:

  • Na₂CO₃: 1 mole* 106 g/mole= 106 g
  • Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
  • CaCO₃: 1 mole* 100 g/mole= 100 g
  • NaNO₃: 2 mole* 85 g/mole= 170 g

You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of  CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of  CaCO₃?

mass of CaCO_{3} =\frac{79.3 grams of Na_{2} CO_{3} *100 grams of of CaCO_{3}}{106 grams of Na_{2} CO_{3}}

mass of CaCO₃= 74.81 grams

<u><em>74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.</em></u>

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A. Electrons have a charge of_____
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Electrons: negative
Protons: positive
Neutrons: nuetral
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