What is the molar mass of NaOCI?

1. 100gm of a 55% (M/M) nitric acid solution is to be diluted to 20% (M/M) nitric acid.

stealth61 [152]

Answer:

The volume of water to be added is 0.175 liters of water

Explanation:

The given concentration of the nitric acid = 55% (M/M)

The mass of the nitric acid solution = 100 gm

The concentration solution is to diluted to = 20% (M/M)

The 100 g 55%(M/M) nitric acid solution gives 55g nitric acid in 100 g of solution

Therefore, to have 20% (M/M) nitric acid solution with the 55 g nitric acid, we get

Let "x" represent the volume of the resulting solution, we have;

20% of x = 55 g of nitric acid

∴ 20/100 × x = 55 g

x = 55 g × 100/20 = 275 g

The mass of extra water to be added = The mass of the 20%(M/M) solution solution of nitric acid - The current mass of the 55%(M/M) solution of nitric acid

The mass of extra water to be added = 275 g - 100 g = 175 g

Volume = Mass/Density

The density of water ≈ 1 g/ml

∴ The volume of water to be added that gives 175 g of water = 175 g/(1 g/ml) = 175 ml. = 0.175 l

The volume of water to be added = 0.175 liters of water.

3 0

3 years ago

using the soubility curve what is the solubilityof nh4cl in 10 mL of water at a temperature of 60 degrees Celsius

lukranit [14]

Answer:

Please, see attached two figures:

The first figure shows the solutility curves for several soluts in water, which is needed to answer the question.

The second figure shows the reading of the solutiblity of NH₄Cl at a temperature of 60°C.

Answer: 5.5g

Explanation:

The red arrow on the second attachement shows how you must go vertically from the temperature of 60ºC on the horizontal axis, up to intersecting curve for the solubility of NH₄Cl.

From there, you must move horizontally to the left (green arrow) to reach the vertical axis and read the solubility: the reading is about in the middle of the marks for 50 and 60 grams of solute per 100 grams of water: that is 55 grams of grams of solute per 100 grams of water.

Assuming density 1.0 g/mol for water, 10 mL of water is:

$10mL\times 1.0g/mL=10g$

Thus, the solutibily is:

$10gWater\times 55gNH_4Cl/100gWater=5.5gNH_4Cl$

5 0

3 years ago

What is the difference between a hydrogen ion and a hydroxide ion?

Sophie [7]

Answer:

there is some difference

Explanation:

hydrogen ion --

symbol-H

Charge- 1+ or 1-

it is atom

hydroxide ion --

symbol- OH

charge-. -1

contains.- one oxygen atom and one hydrogen atom.

they are bonded.

it is a molecule

5 0

3 years ago

What is water's density at 93 ∘C? Assume a constant coefficient of volume expansion. Express your answer with the appropriate un

Ganezh [65]

Answer:

982.5 kg/m³

Explanation:

When the temperature of a fluid increases, it dilates, and because of the variation of the volume, it's density will vary too. The density can be calculated by the expression:

ρ₁ = ρ₀/(1 + β*(t₁ - t₀))

Where ρ₁ is the final density, ρ₀ the initial density, β is the constant coefficient of volume expansion, t₁ the final temperature, and t₀ the initial temperature.

At t₀ = 4°C, the water desity is ρ₀ = 1,000 kg/m³. The value of the constant for water is β = 0.0002 m³/m³ °C, so, for t₁ = 93°C

ρ₁ = 1,000/(1 + 0.0002*(93 - 4))

ρ₁ = 1,000/(1+ 0.0178)

ρ₁ = 982.5 kg/m³

3 0

3 years ago

How many liters of carbon dioxide will 0.5 moles of lithium hydroxide (LiOH) absorb?

Elden [556K]

Answer : The volume of carbon dioxide will be, 5.6122 L

Solution : Given,

Moles of LiOH = 0.5 moles

Molar mass of carbon dioxide = 44 g/mole

Density of carbon dioxide = 0.00196 g/ml

First we have to calculate the mass of carbon dioxide.

The balanced reaction will be,

$2LiOH+CO_2\rightarrow Li_2CO_3+H_2O$

From the reaction we conclude that

As, 2 moles of LiOH absorbs 44 grams of carbon dioxide

So, 0.5 moles of LiOH absorbs $\frac{0.5moles}{2moles}\times 44g=11$ grams of carbon dioxide

Mass of carbon dioxide = 11 g

Density of carbon dioxide = 0.00196 g/ml

Now we have to calculate the volume of carbon dioxide.

$Density=\frac{Mass}{volume}$

$0.00196g/ml=\frac{11g}{volume}$

Volume of carbon dioxide = 5612.24 ml = 5.6122 L (1 L = 1000 ml)

Therefore, the volume of carbon dioxide will be, 5.6122 L

5 0

3 years ago