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Semmy [17]
3 years ago
14

A stock solution of sodium sulfate NaSO4 has a concentrate of 1.00 m. The volume of this solution is 50 ml. What volume of 0.25

M solution could be made from the stock solution
Chemistry
1 answer:
mylen [45]3 years ago
5 0
<h3>Answer:</h3>

200 mL

<h3>Explanation:</h3>

Concept tested: Dilution formula

We are given;

  • Concentration of stock solution as 1.00 M
  • Volume of the stock solution as 50 mL
  • Molarity of the dilute solution as 0.25 M

We are required to calculate the volume of diluted solution;

  • The stock solution is the original solution before dilution while diluted solution is the solution after dilution.
  • Using the dilution formula we can determine the volume of diluted solution;

M1V1 = M2V2

Rearranging the formula;

V2 = M1V1 ÷ M2

     = (1.00 M × 50 mL) ÷ 0.25 M

     = 200 mL

Therefore, a volume of 200mL of 0.25 M solution could be made from the stock solution.

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3 years ago
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Please help I don’t know if my answer is right!!
Temka [501]

Answer:

neutralization reaction! Aka: option C!

HOPE THIS HELPS! :)

Explanation:

8 0
3 years ago
When 70.4 g of benzamide (C7H7NO) are dissolved in 850. g of a certain mystery liquid X, the freezing point of the solution is 2
Arlecino [84]

Answer:

1.62

Explanation:

From the given information:

number of moles of benzamide  =\dfrac{70.4 \ g}{121.14 \ g/mol}

= 0.58 mole

The molality = \dfrac{mass \ of \ solute (i.e. \ benzamide )}{mass \ of \ solvent  }

= \dfrac{0.58 }{0.85 }

= 0.6837

Using the formula:

\mathbf {dT  = l   \times  k_f  \times m}

where;

dT = freezing point = 27

l = Van't Hoff factor = 1

kf = freezing constant of the solvent

∴

2.7 °C = 1 × kf ×  0.6837 m

kf = 2.7 °C/ 0.6837m

kf = 3.949 °C/m

number of moles of NH4Cl = \dfrac{70.4 \ g}{53.491 \  g /mol}

= 1.316 mol

The molality = \dfrac{1.316 \ mol}{0.85 \ kg}

= 1.5484

Thus;

the above kf value is used in determining the  Van't Hoff factor for  NH4Cl

i.e.

9.9 = l × 3.949 × 1.5484 m

l = \dfrac{9.9}{3.949 \times 1.5484 \ m}

l = 1.62

5 0
3 years ago
Position (m)
Alex777 [14]

Option C. The object is returning to the start at a constant speed.

<h3>Data points of the Position vs Time graph</h3>

The following data points will be used to determine the motion of the object.

<u>Position               Time</u>

12                          4

10                          6

2                            8

0                           10

From the data above, the position of the object is decreasing towards zero or start point.

Thus, the object is returning to the start at a constant speed.

Learn more about position here: brainly.com/question/2364404

#SPJ1

8 0
2 years ago
1. Mg2+ and Ca2+ are in the same group on the periodic table. In terms of electronic
finlep [7]

Answer:

Se detailed explanation.

Explanation:

Hello,

In this case, since both magnesium and calcium ions are in group IIA, we can review the following similar properties:

- Since both calcium and magnesium are in group IIA they have two valence electrons, it means that the both of them have two electrons at their outer shells.

- They are highly soluble in water when forming ionic bonds with nonmetals such as those belonging to halogens and oxygen's family.

- Calcium has 18 electrons and magnesium 10 which are two less than the total protons (20 and 12 respectively) since the both of them have lost two electrons due their ionized form.

- Their electron configurations are:

Ca^{20}=1s^2,2s^2,2p^6,3s^3,3p^6,4s^2\\\\Mg^{12}=1s^2,2s^2,2p^6,3s^2

It means that the both of them are at the s region since it is the last subshell at which their electrons are.

Best regards.

8 0
4 years ago
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