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Alexxx [7]
3 years ago
8

Plz answer will give BRAINLIEST! Promise! Question is in picture! Due in 5 minutes!!

Chemistry
1 answer:
zvonat [6]3 years ago
4 0

Answer:

try A

Explanation:

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A stock solution has a concentration of 1.5 M NaCl and is diluted to a 0.80 M solution with a volume of 0.10 L. What volume of t
ioda

Answer:

0.053 L  is the volume of concentrated solution that was used

Explanation:

Let's determine the answer of this, by rules of three.

There is also a dilution formula.

Molarity is a sort of concentration that indicates the moles of solute in 1L of solution.

In 1 L of concentrated solution, there are 1.5 moles of NaCl

In 1 L of diluted solution, there are 0.80 moles.

The volume for the diluted solution is 0.10L

The rule of three will be:

1L of solution has 0.80 moles of solute

Then, 0.10L of solution must have (0.1 . 0.8)/1 = 0.08 moles

This moles came from the concentrated solution, and we know that in 1L of this solution we have 1.5 moles. Therefore the rule of three will be:

1.5 moles are in 1L of solution

0.08 moles were in (0.08 . 1L / 1.5) = 0.053 L (This is the volume of concentrated solution that was used)

Dilution formula is: M conc . Vol conc = M diluted . Vol diluted

1.5 M . Vol conc = 0.80 M . 0.10L

Vol conc = 0.80 M . 0.10L / 1.5M = 0.053L

4 0
3 years ago
Read 2 more answers
24. What volume of a 0.0200M calcium hydroxide is required to neutralize 35.00 mL of 0.0500M nitric acid
Natalka [10]

Answer:

THE VOLUME OF 0.200M CALCIUM HYDROXIDE NEEDED TO NEUTRALIZE 35 mL of 0.050 M NITRIC ACID IS 43.75 mL.

Explanation:

Using

Ca VA / Cb Vb = Na / Nb

Ca = 0.0500 M

Va = 35 mL

Cb = 0.0200 M

Vb = unknown

Na = 2

Nb = 1

Equation for the reaction:

Ca(OH)2 + 2HNO3 --------> Ca(NO3)2 + 2H2O

So therefore, we make Vb the subject of the equation and solve for it

Vb = Ca Va Nb / Cb Na

Vb = 0.0500 * 35 * 1 / 0.0200 * 2

Vb = 1.75 / 0.04

Vb = 43.75 mL

The volume of 0.02M calcium hydroxide required to neutralize 35 mL of 0.05 M nitric acid is 43.75 mL

6 0
3 years ago
Which is the limiting reagent in the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? Reaction: Na2
NemiM [27]

Answer:

CuSO4

Explanation:

Na2S + CuSO4 → Na2SO4 + CuS

The reaction is balanced (same number of elements in each side)

To determine limiting reagent you need to know the moles you have of each.

Molar mass Na2S = 23 * 2 + 32 = 78

Molar mass CuSO4 = 63.5 + 32 + 16 * 4 = 159.5

Na2S mole = 15.5 / 78 = 0.2

CuSO4 mole = 12.1/159.5 = 0.076

*Remember mole = mass / MM

With that information now you have to divide each moles by its respective stoichiometric coefficient

Na2S stoichiometric coefficient : 1

Na2S : 0.2 / 1 = 0.2

CuSO4 stoichiometric coefficient: 1

CuSO4: 0.076 / 1 = 0.076

The smaller number between them its the limiting reagent, CuSO4

8 0
3 years ago
Which test tubes were used to determine the optimal ph for lipase activity?
d1i1m1o1n [39]
The tubes 1, 5 and 6 is the answer to the question which test tubes were used to determine the optimal ph lipase activity. Lipase is an enzyme and optimum ph the maximum possible point on which enzyme become active.
5 0
3 years ago
Because the moon travels around Earth,it is a an blank
Lunna [17]
The answer could be a satellite
6 0
3 years ago
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