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DedPeter [7]
3 years ago
12

PLEASE ANYONE HELP!!!

Chemistry
2 answers:
Fofino [41]3 years ago
7 0
The boiling point increases as more -CH2 groups are added
emmasim [6.3K]3 years ago
6 0

Answer:

I believe it's a.

Explanation:

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Which of the following statements are true? (There are 2)
Shalnov [3]

Answer:

A and B

Explanation:

the other two make no sense at all

4 0
3 years ago
Read 2 more answers
When Z-4,5-dimethyloct-4-ene is treated with hydrogen chloride, HCl, the result is:________.
Vika [28.1K]

Answer:

The  correct option is  c

Explanation:

The chemical equation for the reaction of  Z-4,5-dimethyloct-4-ene and HCl is shown on the first uploaded image

Now looking at the product we see that there are two who has four different groups attached to them this carbon are known as chiral carbons hence the product formed is a pair of diastereomers

6 0
4 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>

6 0
3 years ago
A chemist adds 1.80L of a 1.1/molL aluminum chloride AlCl3 solution to a reaction flask. Calculate the millimoles of aluminum ch
amm1812

Answer:

2000 millimoles of AlCl₃

Explanation:

From the question given above, the following data were obtained:

Volume of solution = 1.8 L

Molarity of solution = 1.1 mol /L

Millmole of AlCl₃ =?

Next, we shall determine the number of mole of AlCl₃ in the solution.

This can be obtained as follow:

Volume of solution = 1.8 L

Molarity of AlCl₃ solution = 1.1 mol /L

Number of mole of AlCl₃ =?

Molarity = mole /Volume

1.1 = Number of mole of AlCl₃ / 1.8

Cross multiply

Number of mole of AlCl₃ = 1.1 × 1.8

Number of mole of AlCl₃ = 1.98 moles

Finally, we shall convert 1.98 moles to millimoles. This can be obtained as follow:

1 mole = 1000 millimoles

Therefore,

1.98 mole = 1.98 mole × 1000 millimoles / 1 mole

1.98 mole = 1980 millimoles

1.98 mole ≈ 2000 millimoles

Thus, the chemist added 2000 millimoles of AlCl₃

7 0
3 years ago
Classify the chemical reaction shown here: CrF3+H3PO4→CrPO4+3HF
Lana71 [14]
The answer 
<span>he chemical reaction CrF3+H3PO4 → CrPO4 + 3HF
the main formula is 
when we observe a reaction as follow:
XY  +  ZT --------------- XT + YZ  
this is classified as double replacement reactions, this means: </span>two ionic compounds exchange ions, producing 2 new ionic compounds.
in our case we have CrF3+H3PO4 → CrPO4 + 3HF
so the classification is double-replacement reactions
5 0
3 years ago
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