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Marianna [84]
3 years ago
6

If a object has a density of 0.55 g/mL , what is it density in cg/L

Chemistry
2 answers:
chubhunter [2.5K]3 years ago
5 0

Answer : The density of a object in cg/L is, 55000 cg/L

Explanation :

As we are given that the density of an object is, 0.55 g/ml.

We have to convert g/ml into cg/L.

As, 1 gram = 100 centigram

As, 1000 milliliter = 1 liter

0.55 g/ml = \frac{0.55g}{1ml}\times \frac{100cg}{1g}\times \frac{1000ml}{1L}=55000cg/L

Therefore, the density of a object in cg/L is, 55000 cg/L

Tpy6a [65]3 years ago
3 0

Answer : The density in cg/L is 0.55\times 10^5cg/L.

Solution :

Conversions :    1g/ml=10^{5}cg/L

Conversion of 'g' to 'cg' :    1g=100cg

Conversion of 'L' to 'ml' :    1L=1000ml

Given density = 0.55g/ml

After conversion,

Density = 0.55\times 10^5cg/L

Therefore, the density in cg/L is 0.55\times 10^5cg/L.


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7 0
2 years ago
Homeostasis is
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4 0
3 years ago
The concentrated sulfuric acid we use in the laboratory is 98.0% sulfuric acid by weight. Calculate the molality and molarity of
timama [110]

Answer : The molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

Solution : Given,

Density of solution = 1.83g/cm^3=1.83g/ml

Molar mass of sulfuric acid (solute) = 98.079 g/mole

98.0 % sulfuric acid by mass means that 98.0 gram of sulfuric acid is present in 100 g of solution.

Mass of sulfuric acid (solute) = 98.0 g

Mass of solution = 100 g

Mass of solvent = Mass of solution - Mass of solute = 100 - 98.0 = 2 g

First we have to calculate the volume of solution.

\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{100g}{1.83g/ml}=54.64ml

Now we have to calculate the molarity of solution.

Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution}}=\frac{98.0g\times 1000}{98.079g/mole\times 54.64ml}=18.29mole/L

Now we have to calculate the molality of the solution.

Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}=\frac{98.0g\times 1000}{98.079g/mole\times 2g}=499.59mole/Kg

Therefore, the molarity and molality of the solution is, 18.29 mole/L and 499.59 mole/Kg respectively.

7 0
3 years ago
Calculate the ph at of a solution of sodium hypochlorite . note that hypochlorous acid is a weak acid with a of . round your ans
loris [4]

The question is incomplete, here is the complete question:

Calculate the pH at 25°C of a 0.39 M solution of sodium hypochlorite NaClO. Note that hypochlorous acid HClO is a weak acid with a pKa of 7.50. Round your answer to 1 decimal place.

<u>Answer:</u> The pH of the solution is 10.4

<u>Explanation:</u>

We are given:

Molarity of sodium hypochlorite = 0.39 M

pK_a of HClO = 7.50

We know that:

pK_a=-\log K_a

K_a  of HClO = 10^{-7.50}=3.16\times 10^{-8}

To calculate the base dissociation constant for the given acid dissociation constant, we use the equation:

K_w=K_b\times K_a

where,

K_w = Ionic product of water = 10^{-14}

K_a = Acid dissociation constant  = 3.16\times 10^{-8}

K_b = Base dissociation constant

Putting values in above equation, we get:

10^{-14}=3.16\times 10^{-8}\times K_b\\\\K_b=\frac{10^{-14}}{3.16\times 10^{-8}}=3.16\times 10^{-7}

The chemical equation for the reaction of hypochlorite ion with water follows:

                    ClO^-+H_2O\rightarrow HClO+OH^-

<u>Initial:</u>           0.39

<u>At eqllm:</u>      0.39-x                   x           x

The expression of K_b for above equation follows:

K_b=\frac{[HClO][OH^-]}{[ClO^-]}

Putting values in above equation, we get:

3.16\times 10^{-7}=\frac{x\times x}{(0.39-x)}\\\\x=-0.00035,0.00035

Neglecting the negative value of 'x' because concentration cannot be negative

To calculate the pOH of the solution, we use the equation:

pOH=-\log[OH^-]

We are given:

[OH^-]=0.00035M

Putting values in above equation, we get:

pOH=-\log (0.00035)=3.6

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH=14-3.6=10.4

Hence, the pH of the solution is 10.4

3 0
3 years ago
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