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3 years ago

What’s the number 9.1x10^3 in standard form

Chemistry

1 answer:

mafiozo [28]3 years ago

3 0

Answer:

.0091

Explanation:

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A gas made up of atoms escapes through a pinhole times as fast as gas. Write the chemical formula of the gas.

Artemon [7]

A gas made up of atoms escapes through a pinhole 0.225times as fast as gas. Write the chemical formula of the gas.

Answer:

Explanation:

To solve this problem, we must apply Graham's law of diffusion. This law states that "the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molecular mass at constant temperature and pressure".

Mathematically;

$\frac{r_{1} }{r_{2} } = \frac{\sqrt{m_{2} } }{\sqrt{m_{1} } }$

r₁ is the rate of diffusion of gas 1

r₂ is the rate of diffusion of gas 2

m₁ is the molar mass of gas 1

m₂ is the molar mass of gas 2

let gas 2 be the given H₂;

molar mass of H₂ = 2 x 1 = 2gmol⁻¹

rate of diffusion is 0.225;

i .e r1/r2 = 0.225

0.225 = √2 / √ m₁

0.225 = 1.414 / √ m₁

√ m₁ = 6.3

m₁ = 6.3² = 39.5g/mol

The gas is likely Argon since argon has similar molecular mass

5 0

3 years ago

What factors trigger the evolution of a species?

lozanna [386]

Humans opinions of evolution.

7 0

3 years ago

Read 2 more answers

Determine the molarity for each of the following solution solutions:

____ [38]

Answer :

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c)The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d)The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e)The molarity of $Br_2$ solution is, 0.0565 mole/L

(f)The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

Explanation :

(a) 1.457 mol of KCl in 1.500 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Solute is KCl.

$\text{Molarity of the solution}=\frac{1.457mole}{1.500L}=0.9713mole/L$

The molarity of KCl solution is, 0.9713 mole/L

(b) 0.515 gram of $H_2SO_4$ , in 1.00 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $H_2SO_4$

Molar mass of $H_2SO_4$ = 98 g/mole

$\text{Molarity of the solution}=\frac{0.515g}{98g/mole\times 1.00L}=0.00525mole/L$

The molarity of $H_2SO_4$ solution is, 0.00525 mole/L

(c) 20.54 g of $Al(NO_3)_3$ in 1575 mL of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $Al(NO_3)_3$

Molar mass of $Al(NO_3)_3$ = 213 g/mole

$\text{Molarity of the solution}=\frac{20.54g\times 1000}{213g/mole\times 1575L}=0.0612mole/L$

The molarity of $Al(NO_3)_3$ solution is, 0.0612 mole/L

(d) 2.76 kg of $CuSO_4.5H_2O$ in 1.45 L of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Solute is $CuSO_4.5H_2O$

Molar mass of $CuSO_4.5H_2O$ = 250 g/mole

$\text{Molarity of the solution}=\frac{2760g}{250g/mole\times 1.45L}=7.61mole/L$

The molarity of $CuSO_4.5H_2O$ solution is, 7.61 mole/L

(e) 0.005653 mol of $Br_2$ in 10.00 ml of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Solute is $Br_2$ .

$\text{Molarity of the solution}=\frac{0.005653mole\times 1000}{10.00L}=0.0565mole/L$

The molarity of $Br_2$ solution is, 0.0565 mole/L

(f) 0.000889 g of glycine, $C_2H_5NO_2$ , in 1.05 mL of solution

Formula used :

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Solute is $C_2H_5NO_2$

Molar mass of $C_2H_5NO_2$ = 75 g/mole

$\text{Molarity of the solution}=\frac{0.000889g\times 1000}{75g/mole\times 1.05L}=0.0113mole/L$

The molarity of $C_2H_5NO_2$ solution is, 0.0113 mole/L

5 0

3 years ago

Traspasa el aire todo hasta llegar a la más alta esfera, y oye allí otro modo de no perecedera música, que es la fuente y la pri

kakasveta [241]

No entiendo cual es tu pregunta?

3 0

3 years ago

3. Determine the moles of sodium, Na, containing 7.9x1024 atoms.

-BARSIC- [3]

Answer:

12.7mol Na.

Explanation:

Hello there!

In this case, according to the concept of mole, which stands for the amount of substance, we can recall the concept of Avogadro's number whereby we understand that one mole of any substance contains 6.022x10²³ particles, for the given atoms of sodium, we can calculate the moles as shown below:

$7.9x10^{23}atoms*\frac{1mol}{6.022x10^{23}atoms} \\\\$

Thus, by performing the division we obtain:

$12.7molNa$

Regards!

8 0

3 years ago