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g100num [7]
3 years ago
6

1. Mixing is a physical change. True or false

Chemistry
2 answers:
Arada [10]3 years ago
6 0

Answer:

false it is a chemical change

Hitman42 [59]3 years ago
5 0

Answer:

true

Explanation:

if u mix it can explode which is a chemical reaction

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When you combine 50.0 mL of 0.100 M AgNO3 with 50.0 mL of 0.100 M HCl in a coffee-cup calorimeter, the temperature changes from
RideAnS [48]

Answer : The enthalpy of reaction (\Delta H_{rxn}) is, 67.716 KJ/mole

Explanation :

First we have to calculate the moles of AgNO_3 and HCl.

\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole

\text{Moles of }HCl=\text{Molarity of }HCl\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole

Now we have to calculate the moles of AgCl formed.

The balanced chemical reaction will be,

AgNO_3(aq)+HCl(aq)\rightarrow AgCl(s)+HNO_3(aq)

As, 1 mole of AgNO_3 react with 1 mole of HCl to give 1 mole of AgCl

So, 0.005 mole of AgNO_3 react with 0.005 mole of HCl to give 1 mole of AgCl

The moles of AgCl formed  = 0.005 mole

Total volume of the solution = 50.0 ml + 50.0 ml = 100.0 ml

Now we have to calculate the mass of solution.

Mass of the solution = Density of the solution × Volume of the solution

Mass of the solution = 1.00 g/ml × 100.0 ml = 100 g

Now we have to calculate the heat.

q=m\times C\Delta T=m\times C \times (T_2-T_1)

where,

q = heat

C = specific heat capacity = 4.18J/g^oC

m = mass = 100 g

T_2 = final temperature = 24.21^oC

T_1 = initial temperature = 23.40^oC

Now put all the given values in the above expression, we get:

q=100g\times (4.18J/g^oC)\times (24.21-23.40)^oC

q=338.58J

Now  we have to calculate the enthalpy of the reaction.

\Delta H_{rxn}=\frac{q}{n}

where,

\Delta H_{rxn} = enthalpy of reaction = ?

q = heat of reaction = 338.58 J

n = moles of reaction = 0.005 mole

Now put all the given values in above expression, we get:

\Delta H_{rxn}=\frac{338.58J}{0.005mole}=6771.6J/mole=67.716KJ/mole

Conversion used : (1 KJ = 1000 J)

Therefore, the enthalpy of reaction (\Delta H_{rxn}) is, 67.716 KJ/mole

4 0
3 years ago
After distilling your crude methyl benzoate, you set aside 4.83 grams of the purified ester. You then prepare the grignard reage
Ber [7]

Answer:

95.6 %

Explanation:

For this question, we will have <u>2 reactions</u>, the formation of the <u>grignard reagent</u> and the <u>formation of the alcohol</u>. The first step then is the calculation of the <u>maximum amount</u> of the grignard reagent. For this, we have to convert the grams to moles and check the smallest value. To do this we have to take into account the <u>following conversion ratios</u>:

Molar mass of Mg = 24 g/mol

Molar mass of phenylmagnesium bromide (C_6H_5Br)= 157 g/mol

Density of bromobenzene= 1.5 g/mL

Molar ratio between Mg and  C_6H_5Br= 1:1

2.3~g~Mg\frac{1~mol~Mg}{24~g~Mg}=0.096~mol~Mg

9.45~mL~\frac{1g}{1.5mL}\frac{1~mol~C_6H_5Br}{157~g}=0.0402~mol~C_6H_5Br

The smallest value is the mol of bromobenzene therefore <u>0.0402 mol</u> of phenylmagnesium bromide would be produced.

The next step is repite the same steps for the reaction of <u>formation of the alcohol</u>. Therefore we have to find the moles of methyl benzoate, so:

Molar mass of methyl benzoate: 136.14 g/mol

4.83~g~\frac{1~mol}{136.14~g}=0.35~mol

The we have to <u>divide by the coefficient</u> of each reactive in the balance reaction. So:

\frac{0.35~mol}{1}=0.35

\frac{0.0402~mol}{2}=0.0201

Therefore the <u>limiting reagent</u> would be the phenylmagnesium bromide. Now, the <u>molar ratio</u> between the phenylmagnesium bromide and triphenyl carbinol is <u>2:1</u>, so the amount of alcohol produced is 0.0201 mol triphenyl carbinol. The next step is the conversion from mol to <u>grams of triphenyl carbinol</u>:

Molar mass of triphenyl carbinol= 260.33 g/mol

0.0201~mol\frac{260.33~g}{1~mol}=5.23~g~triphenyl carbinol

Finally, we have to <u>divide</u> the obtanied solid by the calculated one:

Percentage=\frac{5}{5.23}*100=95.6\%

5 0
3 years ago
Starting with lead (ii) carbonate describe how a solid sample of lead (ii) sulphate can be prepared
tatuchka [14]

The preparation of lead (ii) sulphate from lead (ii) carbonate occurs in two steps:

  1. insoluble lead carbonate is converted to soluble lead (ii) nitrate
  2. soluble lead (ii) nitrate is reacted with sulphuric acid to produce lead (ii) sulphate.

<h3>How can a  solid sample of lead (ii) sulphate be prepared from lead (ii) carbonate?</h3>

Lead (ii) carbonate and  lead (ii) sulphate are both insoluble salts of lead.

In order to prepare lead (ii) sulphate, a two step process is performed.

In the first step, Lead (ii) carbonate is reacted with dilute trioxonitrate (v) acid to produce lead (ii) nitrate.

  • PbCO₃ + 2HNO₃ → Pb(NO₃)₂ + CO₂ + H₂O

In the second step, dilute sulfuric acid is reacted with the lead (ii) nitrate to produce insoluble lead (ii) sulphate which is filtered and dried.

  • Pb(NO₃)₂ + H₂SO₄  →  PbSO₄ + 2HNO₃

In conclusion, lead (ii) sulphate is prepared in two steps.

Learn  more about lead (ii) sulphate at: brainly.com/question/188055

#SPJ1

4 0
2 years ago
Order the terms according to the path followed by oxygen during cellular respiration.​
MArishka [77]

Answer:

Outside air

Nose

Lungs

Bloodstream

Cell

Explanation:

We breathe in oxygen from the outside air in through our nose and it travels to our lungs. Inside our lungs, we have Avioli's that diffuse oxygen into our bloodstream and the bloodstream helps the oxygen travel into our cells.

Hope this helps :)

8 0
3 years ago
In animal cells, DNA is organized in the _______.
Luda [366]
The answer is: Nucleus (same as in plant cells)
5 0
4 years ago
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