What is the molar mass of al (clo2)3

pantera1 [17]

Homogeneous

Hope this helps!!!

6 0

1 year ago

22. How many atoms are there in 344.75 g of gold nugget? a. 1.05 x 10 to the power of 24 atoms b. 1.05 x 10 to the power of 23 a

mixas84 [53]

Answer:

1.053×10²⁴ atoms of gold

Explanation:

Hello,

Gold nugget are usually the natural occurring gold and they contain 85% - 90% weight of pure gold.

In this question, we're required to find the number of atoms in 344.75g of a gold nugget.

We can use mole concept relationship between Avogadro's number and molar mass.

1 mole = molar mass

Molar mass of gold = 197 g/mol

1 mole = Avogadro's number = 6.022 × 10²³ atoms

Number of mole = mass / molar mass

Mass = number of mole × molar mass

Mass = 1 × 197

Mass = 197g

197g is present in 6.022×10²³ atoms

344.75g will contain x atoms

x = (344.75 × 6.022×10²³) / 197

X = 1.053×10²⁴ atoms

Therefore 344.75g of gold nugget will contain 1.053×10²⁴ atoms of gold

5 0

2 years ago

Explain how manipulation of light waves can cause reflection, refraction, diffusion, and absorption; and describe how different

Effectus [21]

Answer:is A

Explanation:

6 0

2 years ago

11. A sample contains 25% water and weighs 201 grams. Determine the grams of water in the

77julia77 [94]

Answer:

$m_{water}=50.25g$

Explanation:

Hello,

In this case, considering that the by-mass percent of water is:

$\% m/m=\frac{m_{water}}{m_{sample}}*100\%$

Given such percent and the mass of the sample, we can find the mass of water in grams in the sample by solving for it as shown below:

$m_{water}=\frac{\%m/m*m_{sample}}{100\%}\\ \\m_{water}=\frac{25\%*201g}{100\%}\\ \\m_{water}=50.25g$

Best regards.

5 0

2 years ago

Suppose you wanted to make a buffer of exactly ph 7.00 using kh2po4 and na2hpo4. if the final solution was 0.10 m in kh2po4, wha

OleMash [197]

Answer:- 0.138 M

Solution:- The buffer pH is calculated using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

$KH_2PO_4$ acts as a weak acid and $Na_2HPO_4$ as a base which is pretty conjugate base of the weak acid we have.

The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:

$H_2PO_4^-\rightleftharpoons H^++HPO_4^-^2$

Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.

So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.

Let's plug in the values in the equation:

$7.00=6.86+log(\frac{base}{0.10})$

$7.00-6.86=log(\frac{base}{0.10})$

$0.14=log(\frac{base}{0.10})$

Taking antilog:

$10^0^.^1^4=\frac{base}{0.10}$

$1.38=\frac{base}{0.10}$

On cross multiply:

[base] = 1.38(0.10)

[base] = 0.138

So, the concentration of the base that is $Na_2HPO_4$ required to make the buffer is 0.138M.

5 0

2 years ago