All categories

2 years ago

Help please need help need help need help for chemistry

Chemistry

1 answer:

Varvara68 [4.7K]2 years ago

8 0

If you type in the letters (uppercase and lowercase matters) into google it will give you the answers

You might be interested in

Glycolic acid, which is a monoprotic acid and a constituent in sugar cane, has a pKa of 3.9. A 25.0 mL solution of glycolic acid

slava [35]

Answer:

pH of resulting solution = 7.98

Explanation:

The balanced equation

HA + NaOH - Na+ + A- + H2O

Number of moles of A = Number of moles of HA = Number of moles of NaOH

= 35.8/1000 * 0.020 = 0.000716 mol

Initial concentration of A = 0.000716/0.0608 = 0.01178 M

pKb = 14 – pKa = 14 -3.9 = 10.1

Kb = 10^{-Kb} = 10^{-10.1} = 7.943 * 10^-11

Kb = [HA][OH-]/[A-]

Kb = a^2/(0.01178 -a) = 7.943 * 10^-11

a^2 + 7.943 * 10^-11 a – 9.357 * 10^-13 = 0

a = 9.673 * 10^-7

OH- = a = 9.673 * 10^-7 M

pOH = -log [OH-] = -log (9.673 * 10^-7) = 6.02

pH = 14-6.02 = 7.98

8 0

2 years ago

How many grams of NaCl (molecular weight = 58.4 g mole-1) would you dissolve in water to make a total volume of 500 mL of soluti

kolezko [41]

Given :

Volume , V = 500 mL .

Molarity , M = 0.5 M .

Molecular mass of NaCl is $58.4\ g/mole$ .

To Find :

How many grams of NaCl is required .

Solution :

Let , NaCl required is x gram .

Molarity is given by :

$M=\dfrac{\text{Number of moles}}{\text{Volume (in liters) }}\\\\M=\dfrac{m}{M\times V}\\\\0.5=\dfrac{x}{58.4\times 0.5}\\\\x=0.5^2\times 58.4\ g\\\\x=14.6\ g$

Hence , this is the required solution.

6 0

2 years ago

A family of four lives in a three-bedroom house and uses an average of 900 kWh of electricity per month. The family cools their

agasfer [191]

<h2>Answer:</h2><h2>The percentage of the family’s total annual electricity that is used to run the two air conditioners for the three summer months = 19.4 %</h2>

Explanation:

Average electricity consumed per month = 900 kWh

The family cools their house for three months during the summer with two window-unit air conditioners

The power consumed by one window-unit air conditioners = 350 kWh

The power consumed by two window-unit air conditioners = 350(2) = 700 kWh

Power consumed for two air conditioners for the three summer months = 700 (3) = 2100 kWh

Total power consumed for 1 year = 900 (12) = 10800kWh

The percentage of the family’s total annual electricity that is used to run the two air conditioners for the three summer months = $\frac{2100}{10800}100$ = 19.4 %