Where is the Sun located?

300.0 mL of a 0.335 M solution of NaI is diluted to 700.0 mL. What is the new concentration of the solution?

Reptile [31]

Answer: The new concentration of the solution is 0.143 M.

Explanation:

Given: $V_{1}$ = 300.0 mL, $M_{1}$ = 0.335 M

$V_{2}$ = 700.0 mL, $M_{2}$ = ?

Formula used is as follows.

$M_{1}V_{1} = M_{2}V_{2}$

Substitute values into the above formula as follows.

$M_{1}V_{1} = M_{2}V_{2}\\0.335 M \times 300.0 mL = M_{2} \times 700.0 mL\\M_{2} = 0.143 M$

Thus, we can conclude that the new concentration of the solution is 0.143 M.

8 0

3 years ago

A mixture of 1.20 mols He, 2.40 mols Ne, 4.80 mols Kr, and 0.60 mols Ar has a total pressure of 600.0 mm Hg. What is the partial

Juliette [100K]

Answer: The partial pressure of the Kr is 320 mm Hg.

Explanation:

According to Raoult's Law , the partial pressure of each component in the solution is equal to the total pressure multiplied by its mole fraction. It is mathematically expressed as

$p_A=x_A\times P_{total}$

where, $p_A$ = partial pressure of component A

$x_A$ = mole fraction of A

$P_{total}$ = total pressure

mole fraction of Krypton = $\frac{\text {Moles of Kr}}{\text {total moles}}=\frac{4.80}{1.20+2.40+4.80+0.60}=0.53$

$p_{Kr}=0.53\times 600mmHg=320mmHg$

Thus partial pressure of the Kr is 320 mm Hg

8 0

3 years ago

3.15.8 cm3 of H2SO4 completely neutralised 20 cm3 of NaOH. The NaOH was 1.5 mol/dm3.

Furkat [3]

Answer:

Molarity = 0.95 mol/dm³

Explanation:

Given data:

Volume of H₂SO₄ = 15.8 cm³

Volume of NaOH = 20 cm³

Concentration of NaOH = 1.5 mol/dm³

Concentration of H₂SO₄ = ?

Solution:

Chemical equation:

NaOH + H₂SO₄ → Na₂SO₄ + H₂O

First of all we will calculate the number of moles of NaOH and for that we will convert the units first,

Volume = 20 cm³/1000 = 0.02 L

Concentration of NaOH = 1.5 mol/dm³

1 mol/dm³ = 1 mol/L

Concentration of NaOH = 1.5 mol/L

Number of moles of NaOH:

Molarity = number of moles / volume in L

1.5 M = number of moles / 0.02 L

Number of moles = 1.5 M ×0.02 L

Number of moles = 0.03 mol

Now we will compare the moles of NaOH and H₂SO₄

NaOH : H₂SO₄

2 : 1

0.03 : 1/2×0.03 = 0.015 mol

Concentration of H₂SO₄:

Volume of H₂SO₄:

15.8 cm³/1000 = 0.0158 L

Molarity = number of moles / volume in L

Molarity = 0.015 mol / 0.0158 L

Molarity = 0.95 mol/L

1 mol/L = 1 mol/dm³

Molarity = 0.95 mol/dm³

6 0

3 years ago

What is Ky for CH3NH2(aq) + H20(1) = CH3NH3(aq) + OH(aq)?

zalisa [80]

I do not have any clue

6 0

3 years ago

The reaction 2HgO (s)→2Hg (I)+O2 (g) has a percent yield of 50%. You want to produce 100 g of Hg.

nevsk [136]

The mass of HgO needed for the reaction is 216 g

The correct answer to the question is Option C. 216 g

We'll begin by calculating the theoretical yield of Hg.

Actual yield of Hg = 100 g

Percentage yield = 50%

Theoretical yield of Hg =?

Theoretical yield = Actual yield / percentage yield

Theoretical yield = 100 / 50%

Theoretical yield of Hg = 200 g

Finally, we shall determine the mass of HgO needed for the reaction.

2HgO → 2Hg + O₂

Molar mass of HgO = 201 + 16 = 217 g/mol

Mass of HgO from the balanced equation = 2 × 217 = 434 g

Molar mass of Hg = 201 g/mol

Mass of Hg from the balanced equation = 2 × 201 = 402 g

From the balanced equation above,

402 g of Hg were produced from 434 g of HgO.

Therefore

200 g of Hg will be produce by = (200 × 434) / 402 = 216 g of HgO.

Thus, 216 g of HgO is needed for the reaction.

Learn more about stoichiometry:

brainly.com/question/24426334

4 0

2 years ago

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