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dsp73
2 years ago
10

Calculate the volume needed of a stock (undiluted) solution that has a concentration of 4.2 M and is then diluted to 0.034 L of

0.66 M solution.
Chemistry
1 answer:
Mnenie [13.5K]2 years ago
5 0

M1V1 = M2V2

V1(4.2) = (0.66)(0.034)

V1 = 0.0053L

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LenaWriter [7]

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4 0
3 years ago
Also need to know some of this stuff
irinina [24]

Answer:

answers are in attachment

Explanation:

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6 0
3 years ago
C + 2H2 -> CH4
ankoles [38]

Answer:

<u>C) 4</u>

Explanation:

<u>The reaction</u> :

  • C (s) + 2H₂ (g) ⇒ CH₄ (g)

       12g      4g             16g

Hence, based on this we can say that : <u>2 moles of hydrogen gas are needed to produce 16g of methane.</u>

<u />

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4 0
1 year ago
Read 2 more answers
You have a particle of length 68 nm. What is this in meters? ( need help asap)
Bess [88]

Answer:

B

Explanation:

4 0
2 years ago
PLEASE HELP!<br><br> See picture
Darya [45]

Answer:

See Explanation

Explanation:

The equation of the reaction;

KHSO4(aq) + KOH(aq) -------> K2SO4(aq) + H2O(l)

Number of moles of KHSO4 = 49.6 g/136.169 g/mol = 0.36 moles

Since the reaction is in a mole ratio of 1:1, 0.36 moles of K2SO4 is produced.

Number of moles of KOH = 25.3 g/56.1056 g/mol = 0.45 moles

Since the reaction is 1:1, 0.45 moles of K2SO4 is produced

Hence K2SO4 is the limiting reactant.

Mass of K2SO4 formed = 0.36 moles of K2SO4 * 174.26 g/mol = 62.7 g

So;

1 mole of KHSO4 reacts with 1 mole of KOH

0.36 moles of KHSO4 reacts with 0.36 * 1/1 = 0.36 moles of KOH

Amount of excess KOH = 0.45 moles - 0.36 moles = 0.09 moles

Mass of excess KOH = 0.09 moles * 56.1056 g/mol  = 5 g of excess KOH

5 0
3 years ago
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