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oksian1 [2.3K]
2 years ago
10

Which of the following statements are true of groups on the periodic table?

Chemistry
1 answer:
Gnoma [55]2 years ago
6 0

Answer: The correct answer is A, C and D

Explanation:

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Determine the pHpH of an HFHF solution of each of the following concentrations. In which cases can you not make the simplifying
PIT_PIT [208]

The question is incomplete, complete question is :

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (K_a for HF is 6.8\times 10^{-4}.)

[HF] = 0.280 M

Express your answer to two decimal places.

Answer:

The pH of an 0.280 M HF solution is 1.87.

Explanation:3

Initial concentration if HF = c = 0.280 M

Dissociation constant of the HF = K_a=6.8\times 10^{-4}

HF\rightleftharpoons H^++F^-

Initially

c          0            0

At equilibrium :

(c-x)      x             x

The expression of disassociation constant is given as:

K_a=\frac{[H^+][F^-]}{[HF]}

K_a=\frac{x\times x}{(c-x)}

6.8\times 10^{-4}=\frac{x^2}{(0.280 M-x)}

Solving for x, we get:

x = 0.01346 M

So, the concentration of hydrogen ion at equilibrium is :

[H^+]=x=0.01346 M

The pH of the solution is ;

pH=-\log[H^+]=-\log[0.01346 M]=1.87

The pH of an 0.280 M HF solution is 1.87.

6 0
2 years ago
Sodium nitrate is used in fertilizers, explosives, and glass manufacturing. Give the formula for this compound
krek1111 [17]

Explanation:

The molecular formula for Sodium Nitrate is NaNO3. The SI base unit for amount of substance is the mole. 1 grams Sodium Nitrate is equal to 0.011765443644878 mole.

7 0
2 years ago
It takes 90 N of force to lift a box of books off my smiths table. If you can only leave the box 25% of the way up, how many new
leva [86]

Answer:3.6 I think sorry if wrong

Explanation:

90 divided by 25

4 0
2 years ago
A 10​-liter ​[l] flask contains 1.4 moles​ [mol] of an ideal gas at a temperature of 20 degrees celsius ​[degrees​c]. What is th
Lynna [10]

The pressure in the flask is 3.4 atm.

<em>pV</em> = <em>nRT </em>

<em>T</em> = (20 + 273.15) K = 293.15 K

<em>p</em> = (<em>nRT</em>)/<em>V</em> = (1.4 mol × 0.082 06 L·atm·K⁻¹mol⁻¹ × 293.15 K)/10 L = 3.4 atm

4 0
2 years ago
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In a titration experiment a student uses 1.4 m hbr solution and the indicator phenolphthalein to determine the concentration of
Likurg_2 [28]
The question is incomplete. Complete question is attached below:
...........................................................................................................................

Answer: 
Given: conc. of HBr = 1.4 M
Volume of HBr = 15.4 mL
Volume of KOH = 22.10 mL

We know that, M1V1 = M2V2
                        (HBr)      (KOH)

Therefore, M2 = M1V1/V2
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                        = 0.9756 M

Concentration of KOH is 0.9756 M.

3 0
3 years ago
Read 2 more answers
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