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Natasha2012 [34]
2 years ago
6

What volume of 5.00 M KI stock solution is needed to make 250. mL of a 1.50 M solution?

Chemistry
2 answers:
brilliants [131]2 years ago
6 0
So it says 5.00 M KL so we use that number and now we have 250 mL so if we use those to number and the 1.50 we get C
Step2247 [10]2 years ago
5 0

Answer:

B

Explanation:

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Na sodiops los elementos químicos se representan como el átomo, el núcleo, donde se necuentran los protones y neutrones van dentro, los electrones afuera, girando de forma elíptica alrededor del núcleo. ... por ejemplo el oxígeno numero atómico 8, tiene 8 protones y 8 electrones. el número de neutrones es diferente.

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Sodium-potassium pumps are examples of what type of cellular transport?
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3 years ago
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3 years ago
What is the distinction between magma and lava​
lesya692 [45]

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7 0
2 years ago
Imagine you had HCl with a concentration of exactly 0.10 mol/dm3. If 0.023 dm3 of a sodium hydroxide solution, NaOH (aq), could
polet [3.4K]

Answer:

Explanation:

Step 1: Calculate the amount of sodium hydroxide in moles

Volume of sodium hydroxide solution = 25.0 ÷ 1,000 = 0.0250 dm3

Rearrange:

Concentration in mol/dm3 =

Amount of solutein mol = concentration in mol/dm3 × volume in dm3

Amount of sodium hydroxide = 0.100 × 0.0250

= 0.00250 mol

Step 2: Find the amount of hydrochloric acid in moles

The balanced equation is: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

So the mole ratio NaOH:HCl is 1:1

Therefore 0.00250 mol of NaOH reacts with 0.00250 mol of HCl

Step 3: Calculate the concentration of hydrochloric acid in mol/dm3

Volume of hydrochloric acid = 20.00 ÷ 1000 = 0.0200 dm3

Concentration in mol/dm3 =

Concentration in mol/dm3 =

= 0.125 mol/dm3

Step 4: Calculate the concentration of hydrochloric acid in g/dm3

Relative formula mass of HCl = 1 + 35.5 = 36.5

Mass = relative formula mass × amount

Mass of HCl = 36.5 × 0.125

= 4.56 g

So concentration = 4.56 g/dm3

8 0
2 years ago
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