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Lena [83]
3 years ago
9

A machine

Chemistry
2 answers:
rosijanka [135]3 years ago
4 0

i believe the answer is A (((:

AlexFokin [52]3 years ago
4 0

The answer is D)

A machine can do all of these things

Depending on what kind of machine it is, it can change direction and shapes of objects. Machines can usually do these things much faster than a person can.

I hope this Helped :D

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write a balanced equation to determine the molarity of the HCI solution when a 24.6 ml sample of HCI reacts with a 33.0 mL of 0.
MatroZZZ [7]

Answer: 0.30 M

Explanation:

HCl+NaOH\rightarrow NaCl+H_2O

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

M_1 = molarity of HCl solution = ?

V_1 = volume of HCl solution = 24.6 ml

M_2 = molarity of NaOH solution = 0.222 M

V_2 = volume of NaOH solution = 33.0 ml

n_1 = valency of HCl = 21

n_2 = valency of NaOH = 1

1\times M_1\times 24.6=1\times 0.222\times 33.0

M_1=0.30M

Therefore, the molarity of the HCI solution is 0.30 M

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3 years ago
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What is El Nino and La Nina, and how are they different?
gavmur [86]
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3 years ago
A bond where the electrons are shared equally is called a(an) ________ bond.
zubka84 [21]
Metallic bonds result when electrons are shared equally.

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4 years ago
Identify an alkene and carboxylic acid using primary observations
ehidna [41]

Answer:

The general formula for the carboxylic acids is C nH 2n+1COOH (where n is the number of carbon atoms in the molecule, minus 1).

Explanation:

<em>Hope </em><em>it </em><em>helps </em><em>u </em>

FOLLOW MY ACCOUNT PLS PLS

4 0
3 years ago
A student needs to prepare 50.0 mL of 0.80 M aqueous H2O2 solution. Calculate the volume of 4.6 M H2O2 stock solution that shoul
Degger [83]

Answer:

We need 8.7 mL of the stock solution

Explanation:

Step 1: Data given

Volume of the solution he wants to prepare = 50.0 mL = 0.050 L

Concentration of the solution he wants to prepare = 0.80 M

The concentration of the stock solution = 4.6 M

Step 2: Calculate the volume of the stock solution

C1*V1 = C2*V2

⇒with C1 = the concentration of the stock solution = 4.6 M

⇒with V1 = the volume of the stock solution = TO BE DETERMINED

⇒with C2 = the concentration of the prepared solution = 0.80 M

⇒with V2 = the volume of the prepared solution = 0.050 L

4.6 M * V2 = 0.80 M * 0.050 L

V2 = (0.80 M * 0.050 L) / 4.6M

V2 = 0.0087 L = 8.7 mL

We need 8.7 mL of the stock solution

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4 years ago
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