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wlad13 [49]
2 years ago
6

What volume in milliliters of 6.0 M NaOH is needed to prepare 175mL of 0.20 M NaOH by dilution?

Chemistry
1 answer:
Ilya [14]2 years ago
4 0

Answer:

V¹N²= V²N²

here V¹= ?

N¹= 6.00

V²= 175ml

M²= 0.2M

So V¹= (V²N²)/N² = (175 x 0.2)/6

V¹ = 5.83 ml

Explanation:

Therefore diluting 5.83 ml of 6.00M NaOH to 175 m l ,we get 0.2M Solution.

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Consider four sealed, rigid containers with the following volumes: 50 mL, 100 mL, 250 mL, and 500 mL. If each of these contains
dsp73
First, we assume that helium behaves as an ideal gas such that the ideal gas law is applicable.
                                     PV = nRT
where P is pressure, V is volume, n is number of moles, R is universal gas constant, and T is temperature. From the equation, if n, R, and T are constant, there is an inverse relationship between P and V. From the given choices, the container with the greatest pressure would be the 50 mL. 
5 0
2 years ago
The combustion of propane occurs when propane interacts with oxygen gas to produce water and carbon dioxide in the following rea
gtnhenbr [62]

Answer:

  • 20 atm

Explanation:

<u>1) Data:</u>

a) n = 2 moles

b) T = 373 K

c) V = 2.5 liter

d) P = ?

<u>2) Chemical principles and formula</u>

You need to calculate the pressure of the propane gas in the mixture before reacting. So, you can apply the partial pressure principle which states that each gas exerts a pressure as if it occupies the entire volume.

Thus, you just have to use the ideal gas equation: PV = nRT

<u>3) Solution:</u>

  • PV = nRT ⇒ P = nRT / V

P = 2 mol × 0.08206 atm-liter /K-mol × 373K / 2.5 liter = 24.5 atm

Since the number of moles are reported with one significant figure, you must round your answer to one significant figure, and that is 20 atm (20 is closer to 24.5 than to 30).

5 0
2 years ago
How many milligrams of MgI2 must be added to 257.7 mL of 0.087 M KI to produce a solution with [I−] = 0.1000 M?
lbvjy [14]

Answer:

The answer is 465.6 mg of MgI₂ to be added.

Explanation:

We find the mole of ion I⁻ in the final solution

C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol

But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.

So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.

Hence, the weight of MgI₂ must be added is

Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg

4 0
2 years ago
An unknown solid is entirely soluble in water. On addition of dilute HCl, a precipitate forms. After the precipitate is filtered
Karo-lina-s [1.5K]

Answer:

Pb(NO3)2

Cd(NO3)2

Na2SO4

Explanation:

In the first part, addition of HCl leads to the formation of PbCl2 which is poorly soluble in water. This is the first precipitate that is filtered off.

When the pH is adjusted to 1 and H2S is bubbled in, CdS is formed. This is the second precipitate that is filtered off.

After this precipitate has been filtered off and the pH is adjusted to 8, addition of H2S and (NH4)2HPO4 does not lead to the formation of any other precipitate.

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5 0
1 year ago
Which best describes the relationships between subatomic particles in any neutral atom? * 1 point the number of electrons equals
Anna71 [15]

Atomic number = protons

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Atomic number = atomic mass = neutrons

P = E

AM - AN = N

Example:

Calcium = 20 Protons  20P = 20E

Atomic mass - atomic number = neutrons :)

4 0
2 years ago
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