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Veronika [31]
2 years ago
13

Problems that I am stuck on! Need the answers asap

Chemistry
1 answer:
Anna71 [15]2 years ago
6 0

Answer:

1.b

2.b

3.a

4.a

Explanation:

use the rate laws when answering this.

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Aqueous sulfuric acid H2SO4 will react with solid sodium hydroxide NaOH to produce aqueous sodium sulfate Na2SO4 and liquid wate
allochka39001 [22]

Answer:

a) 2NaOH+H2SO4→Na2SO4+2H2O2NaOH+H2SO4→Na2SO4+2H2O

b) Số phân tử NaOH : Số phân tử H2SO4 = 2:1

Số phân tử NaOH : Số phân tử Na2SO4 = 2:1

Số phân tử NaOH : Số phân tử H2O = 2:2

Explanation:

8 0
2 years ago
Give the reason for the following; 1. Iodimetric titrations are usually performed in neutral or mildly alkaline (pH 8) to weakly
Agata [3.3K]

Answer:

Explanation:

blach bal

5 0
3 years ago
How many moles of 0.225 M CaOH2 are present in 0.350 L of solution?
weeeeeb [17]

Answer : The number of moles of solute Ca(OH)_2 is, 0.0788 moles.

Explanation : Given,

Molarity = 0.225 M

Volume of solution = 0.350 L

Formula used:

\text{Molarity}=\frac{\text{Moles of }Ca(OH)_2}{\text{Volume of solution (in L)}}

Now put all the given values in this formula, we get:

0.225M=\frac{\text{Moles of }Ca(OH)_2}{0.350L}

\text{Moles of }Ca(OH)_2=0.0788mol

Therefore, the number of moles of solute Ca(OH)_2 is, 0.0788 moles.

7 0
3 years ago
You need to prepare at 2.0 mL sample of a diluted drug for injection. The total amount of the drug to be injected in this 2.0 mL
pav-90 [236]

Answer:

a) The concentration of drug in the bottle is 9.8 mg/ml

b) 0.15 ml drug solution + 1.85 ml saline.

c) 4.9 × 10⁻⁵ mol/l

Explanation:

Hi there!

a) The concentration of the drug in the bottle is 294 mg/ 30.0 ml = 9.8 mg/ml

b) The drug has to be administrated at a dose of 0.0210 mg/ kg body mass. Then, the total mass of drug that there should be in the injection for a person of 70 kg will be:

0.0210 mg/kg-body mass * 70 kg = 1.47 mg drug.

The volume of solution that contains that mass of drug can be calculated using the value of the concentration calculated in a)

If 9.8 mg of the drug is contained in 1 ml of solution, then 1.47 mg drug will be present in (1.47 mg * 1 ml/ 9.8 mg) 0.15 ml.

To prepare the injection, you should take 0.15 ml of the concentrated drug solution and (2.0 ml - 0.15 ml) 1.85 ml saline

c) In the injection there is a concentration of (1.47 mg / 2.0 ml) 0.735 mg/ml.

Let´s convert it to molarity:

0.735 mg/ml * 1000 ml/l * 0.001 g/mg* 1 mol/ 15000 g = 4.9 × 10⁻⁵ mol/l

3 0
3 years ago
Hewo fellow earthlings
kobusy [5.1K]

Answer:

Hello!

Explanation:

5 0
2 years ago
Read 2 more answers
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