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solniwko [45]
2 years ago
15

Question 3 0 / 1 point How many grams of carbon atoms are present in a sample of C3H8 if there are 4.10 moles of hydrogen atoms

in the sample
Chemistry
1 answer:
hodyreva [135]2 years ago
5 0

Answer:

18.45 g of C

Explanation:

This is a problem of rules of three:

1 mol of C₃H₈ contains 3 moles of C and 8 moles of H

If 8 moles of H are contained in 1 mol of propane

4.10 moles of H are contained in (4.1 . 1) /8 = 0.5125 moles

Now, If 1 mol of propane contains 3 moles of C

0.5125 moles of propane may contain (0.5125 . 3) / 1 = 1.5375 moles of C

Let's convert the moles to mass:

1.5375 mol . 12 g /mol = 18.45 g

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Chemical formula for Potassium Hydrogen Nitrate
My name is Ann [436]

Answer:

KNO3

Explanation:

8 0
3 years ago
Aqueous lithium sulfate was mixed with aqueous strontium chlorate, and a crystallized strontium sulfate product was formed. cons
Dmitriy789 [7]

Lithium sulfate is Li2SO4

Strontium chlorate is Sr(ClO3)2

Strontium sulfate is SrSO4

 

So the complete balanced chemical reaction for this is:

 

Li2SO4 (aq) +  Sr(ClO3)2 (aq) -->  SrSO4 (s)  +  2 LiClO3 (aq)

 

 

This is a type of double replacement reaction since there is an exchange of ions.

4 0
3 years ago
Name the following compound:<br> CH3-CH2-CH2-CH2-CH3<br> CH3 CH3
stepan [7]

Answer:

<u><em>Pentane </em></u>

Explanation:

since we have in here CH3-CH2-CH2-CH2-CH3 5 Carbon atoms and 12 Hydrogen making it C_{5} H_{12}

6 0
2 years ago
8. The standard enthalpy of formation of RbF(s) is –557.7kJ/mol and the standard enthalpy of formation of RbF(aq, 1 m) is –583.8
garri49 [273]

Explanation:

Given

The enthalpy of formation of RbF (s) is –557.7kJ/mol

The standard enthalpy of formation of RbF (aq, 1 m) is –583.8 kJ/mol

The enthalpy of solution of RbF = Enthalpy of RbF (aq) - Enthalpy of formation of RbF (s)

= -583.8 - (-557.7)  kJ/mol

= -26.1 kJ/mol

The enthalpy is negative which means that the temperature will rise when RbF is dissolved.

3 0
3 years ago
William measures a test tube and finds that the mass of the test tube is 5g. He places the lone reactant in the test tube and fi
iogann1982 [59]

86 percent is the percent yield for this experiment if he expected to produce 5g of product.

Explanation:

Given that:

mass of test tube = 5 grams

mass of test tube + reactant is 12.5 grams

mass of reactant = ( mass of test tube + reactant ) - (mass of test tube)

mass of reactant = 12.5 -5

                             = 7.5 grams

when 7.5 grams of reactant is heated mass of test tube was found to be 9.3 grams.

so mass of product formed = 9.3 - 5

                                             = 4. 3 grams of product is formed (actual yield)

However, he expected the product to be 5 grams (theoretical yield)

Percent yield = \frac{actual yield}{theoretical yield} x 100

          putting the values in the formula:

percent yield = \frac{4.3}{5} x 100

                     = 86 %

86 percent is the percent yield.

5 0
3 years ago
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