From the formula MgO how do you know that Mg is the metal?

You are so excited-your first summer job! You are a hot-tub maintenance assistant. You travel to homes, making sure the pH of th

AfilCa [17]

Answer: well i guess we can say that is a scale used to specify the acidity or basicity of an aqueous solution and it is important to keep ph of hot tub water between 7 and 7.6 because if it is not balanced the water can negatively affect your tub.

Explanation:

pH is an extremely important chemical level to maintain in your spa's water chemistry. A high pH or a low pH will spell trouble for you and your spa. On the pH scale, a reading of 7.4-7.6 indicates perfectly balanced pH. Any number lower than 7.4 means that your water is now considered “acidic”.

6 0

3 years ago

Read 2 more answers

Antacids are taken to counterbalance excess hydrochloric acid, HCl, in the stomach. Complete and balance the following neutraliz

abruzzese [7]

Answer:

1. CaCO3 + 2HCl —> CaCl2 + H2O + CO2

2. Mg(OH)2 + 2HCl —> MgCl2 + 2H2O

3. Na2CO3 + 2HCl —> 2NaCl + H2O + CO2

7 0

3 years ago

What is the molar mass of potassium bromide, KBR?

Nezavi [6.7K]

Answer:

id.............k

Explanation:

3 0

3 years ago

a sample of CO2 occupies a volume of 280ml at a pressure of 1.3 atm and a temperature of 18 degrees celsius, what volume will th

monitta

We can calculate the new volume of the gas using the Combined Gas Law:

(P1 x V1) / T1 = (P2 x V2) / T2

The initial volume, pressure, and temperature were 280 mL, 1.3 atm, and 291.15 K (changing the temperature into Kelvin is necessary), and the final volume, pressure, and temperature is V2, 3.0 atm, and 308.15 K. Plugging these values in and solving, we find that:

(P1 x V1) / T1 = (P2 x V2) / T2

(1.3 atm x 280 mL) / 291.15 K = (3.0 atm x V2) / 308.15 K

V2 = 128.42 mL

This makes sense considering the conditions, a small increase in temperature would make the gas expand but a significant increase in the pressure would cause the volume to decrease.

Hope this helps!

7 0

3 years ago

A certain liquid has a normal boiling point of and a boiling point elevation constant . A solution is prepared by dissolving som

jek_recluse [69]

The question is incomplete, the complete question is:

A certain substance X has a normal freezing point of $-6.4^oC$ and a molal freezing point depression constant $K_f=3.96^oC.kg/mol$ . A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at $-13.6^oC$ . Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.

<u>Answer:</u> The mass of glycine that can be dissolved is $1.3\times 10^2g$

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m$

OR

$\text{Freezing point of pure solvent}=\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

where,

Freezing point of pure solvent = $-6.4^oC$

Freezing point of solution = $-13.6^oC$

i = Vant Hoff factor = 1 (for non-electrolytes)

$K_f$ = freezing point depression constant = $3.96^oC/m$

$m_{solute}$ = Given mass of solute (glycine) = ?

$M_{solute}$ = Molar mass of solute (glycine) = 75.07 g/mol

$w_{solvent}$ = Mass of solvent = 950. g

Putting values in equation 1, we get:

$-6.4-(-13.6)=1\times 3.96\times \frac{m_{solute}\times 1000}{75.07\times 950}\\\\m_{solute}=\frac{7.2\times 75.07\times 950}{1\times 3.96\times 1000}\\\\m_{solute}=129.66g=1.3\times 10^2g$

Hence, the mass of glycine that can be dissolved is $1.3\times 10^2g$

5 0

3 years ago