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ICE Princess25 [194]
3 years ago
8

How would you male 2.0L of O.100M sulfurice acid from a 12.0M stock bottle of sulfurice acid Calc, sketch, and write a summary s

tatement
Chemistry
1 answer:
ozzi3 years ago
7 0

<u>Answer:</u> The volume of stock solution needed is 0.016 L

<u>Explanation:</u>

To calculate the volume of stock solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the stock solution

M_2\text{ and }V_2 are the molarity and volume of diluted solution

We are given:

M_1=12M\\V_1=?L\\M_2=0.10M\\V_2=2L

Putting values in above equation, we get:

12\times V_1=0.1\times 2\\\\V_1=0.016L

Hence, the volume of stock solution needed is 0.016 L

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Mass is not conserved in chemical reactions. Mass is therefore never conserved because a little of it turns into energy in every reaction
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From the following balanced equation,
ArbitrLikvidat [17]

Answer:

9.72 grams.

Explanation:

From the equation, 4 moles of NH₃ produce 6 moles of  water.

Therefore the reaction to product ratio of  NH₃ to H₂O is 4:6

and 2:3 into its simplest form.

The number of moles of NH₃ in 6.12 g is:

Number of moles=mass/ RMM

=6.12 g/17 G/mol

=0.36 moles.

Therefore the number of moles of H₂O produced is calculated as follows.

(0.36 Moles×3)2 = 0.54 moles

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6 0
3 years ago
Can Neon form any compounds?
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3 years ago
Write the balanced chemical equation for the combination (synthesis) reaction of magnesium with oxygen gas. phases are optional.
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3 years ago
PLEASE HELP ASAP!!!!
Vsevolod [243]

Answer:

The answer to your question is  C = 0.037 cal/g°C

Explanation:

Data

mass of the wire = m = 237 g

temperature 1 = T1 = 25°C

temperature 2 = T2 = 107°C

Heat = Q = 722 cal

Specific heat = C

Process

1.- Write the formula to find the specific heat

            Q = mC(T2 - T1)

-Solve for C

            C = Q / m(T2 - T1)

2.- Substitution

             C = 722 / 237(107 - 25)

3.- Simplification

              C = 722 / 237(82)

              C = 722 / 19434

4.- Result

               C = 0.037 cal/g°C

4 0
3 years ago
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