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natta225 [31]
2 years ago
11

If you are launching a matchstick rocket, the action force is the rocket pushing the gases down. what is the reaction force

Chemistry
1 answer:
Andru [333]2 years ago
7 0
Its skleet skal skloop boom its in the air jk
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The empirical formula for a compound is CH2. If n is a whole number, which shows a correct relationship between the molecular fo
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Molecular formula = empirical formula * n
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A student noticed a yellow, solid substance was formed when mixing two clear liquids during a lab. What term describes this yell
givi [52]

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D

Explanation:

a precipitate is formed from a solution

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This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): An analytical chemist has det
murzikaleks [220]

Answer:

0.11 mol

Explanation:

<em>This is the chemical formula for acetic acid (the chemical that gives the sharp taste to vinegar): CH₃CO₂H. An analytical chemist has determined by measurements that there are 0.054 moles of oxygen in a sample of acetic acid. How many moles of hydrogen are in the sample?</em>

Step 1: Given data

  • Formula of acetic acid: CH₃CO₂H
  • Moles of oxygen in the sample of acetic acid: 0.054 moles

Step 2: Establish the appropriate molar ratio

According to the chemical formula of acetic acid, the molar ratio of H to O is 4:2.

Step 3: Calculate the moles of atoms of hydrogen

We will use the theoretical molar ratio for acetic acid.

0.054 mol O × (4 mol H/2 mol O) = 0.11 mol H

3 0
2 years ago
Suggest why sodium and hydrogen ions do not diffuse at the same rate
Troyanec [42]

Answer:

sodium has got ionic bonds that are weak

compared to hydrogen covalent bonds that are strong

8 0
2 years ago
What is the pH of a solution prepared by mixing 50.00 mL of 0.10 M methylamine, CH3NH2, with 20.00 mL of 0.10 M methylammonium c
olya-2409 [2.1K]

Answer:

pH=10.97

Explanation:

the solution of methyl amine with methylammonium chloride will make a buffer solution.

The pH of buffer solution can be obtained using Henderson Hassalbalch's equation, which is:

pOH=pKb+log\frac{[salt]}{[base]}

pH = 14- pOH

Let us calculate pOHpOH=3.43+ (-0.397)=3.03

pH=14-pOH=14-3.03=10.97

[Salt] = [methylammonium chloride] = 0.10 M (initial)

After adding base

[salt] = \frac{molarityXvolume}{finalvolume}=\frac{0.1X20}{(20+50)}= 0.0286M

[base] = [Methylamine]=0.10

After mixing with salt

[base]= \frac{molarityXvolume}{finalvolume}=\frac{0.1X50}{(20+50)}= 0.0714M

pKb= -log[Kb]= 3.43

Putting values

pOH = 3.43+log(\frac{[0.0286]}{0.0714}

4 0
2 years ago
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