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3 years ago

How does the structure of Ethane (C2H6) and Propane(C3H8) differ? Are they members of a Homologous series?

Chemistry

1 answer:

aksik [14]3 years ago

5 0

Answer:

Ethane & propane are members of same homologous series : alkane which is indicated with Cn H2n+2

Explanation:

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Is this right????????????

zzz [600]

Answer: yea its right

Explanation:

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3 years ago

Read 2 more answers

1. In order to make 1 salad it requires 1 head of lettuce, two tomatoes and three carrots. What would be the coefficients for th

pogonyaev

Answer:

3 salad = 3 lettuce + 6 tomatoes + 9 three carrots

Coefficients: 3, 6, 9

Explanation:

1 salad = 1 lettuce + 2 tomatoes + 3 three carrots

Multiply all the coefficients of 1 salad by 3:

3(1 salad) = 3(1 lettuce + 2 tomatoes + 3 three carrots)

Expand the equation:

3 salad = 3 lettuce + 6 tomatoes + 9 three carrots

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3 years ago

If you start with 89.3 g no(g) and 28.6 g h2(g), find the theoretical yield of ammonia.

Tatiana [17]

Balanced equation:
2 NO + 5 H2 ------> 2 NH3 + 2 H2O

2 moles NO react with 5 moles H2 to produce 2 moles NH3

Molar mass of NO = 30.00 g/mol 
86.3g NO = 86.3/30.00 = 2.877 moles of NO 

This will require: 2.877*5 / 2 = 7.192 moles of H2 

Molar mass of H2 = 2 g/mol 
25.6g H2 = 25.6/2 = 12.7 mol H2. 
You have excess H2 means the NO is limiting 

From the balanced equation: 
2 moles of NO will produce 2 moles of NH3 
2.877 moles of NO will produce 2.877 moles of NH3 

Molar mass NH3 = 17g/mol 
Mass NH3 produced = 2.877 * 17 = 48.91g

Hence the yield is = 48.91 g ~ 49 g

3 0

3 years ago

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You make the following measurements of an object – 22 kg and 42 m3. What would the object’s density be? Show your work for credi

Dafna11 [192]

Density = mass / volume
therefore, density = 22/42 = 0.524 kg/m3

4 0

3 years ago

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Calculate the maximum volume (in mL) of 0.143 M HCl that each of the following antacid formulations would be expected to neutral

Nuetrik [128]

Answer:

a. The maximum volume of 0.143 M HCl required is 154.4 mL.

b. The maximum volume of 0.143 M HCl required is 135.7 mL.

Explanation:

$Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O$

Mass of aluminum hydroxide = 350 mg = 0.350 g ( 1mg = 0.001 g)

Moles of aluminum hydroxide = $\frac{0.350 g}{78 g/mol}=0.004487 mol$

According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :

$\frac{3}{1}\times 0.004487 mol=0.01346 mol$ of HCl.

$Mg(OH)_2+2HCl\rightarrow MgCL_2+2H_2O$

Mass of magnesium hydroxide = 250 mg = 0.250 g ( 1mg = 0.001 g)

Moles of magnesium hydroxide = $\frac{0.250 g}{58 g/mol}=0.004310 mol$

According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310 mole of magnesium hydroxide will be neutralize by :

$\frac{2}{1}\times 0.004310 mol=0.008621 mol$ of HCl.

Total moles of HCl required to neutralize both :

0.01346 mol + 0.008621 mol = 0.02208 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

$Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}$

$V=\frac{0.02208 mol}{0.143 M}=0.1544 L$

1 L = 1000 mL

0.1544 L = 154.4 mL

The maximum volume of 0.143 M HCl required is 154.4 mL.

$CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2$

Mass of calcium carbonate = 970mg = 0.970 g ( 1mg = 0.001 g)

Moles of calcium carbonate = $\frac{0.970 g}{100 g/mol}=0.00970 mol$

According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :

$\frac{2}{1}\times 0.00970 mol=0.0194 mol$ of HCl.

Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

$Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}$

$V=\frac{0.0194 mol}{0.143 M}=0.1357 L$

1 L = 1000 mL

0.1357 L = 135.7 mL

The maximum volume of 0.143 M HCl required is 135.7 mL.

4 0

3 years ago