What is the molar mass of NH4OH

What is 0.00550 g converted to mg

Mila [183]

Answer: it is 5.5 mg

Explanation:

you have to multiply the mass value by 1000

5 0

3 years ago

Pls hlp me with these questions: brainliest, ratting, thx, etc.

daser333 [38]

1 elements
2 they have same number of valence electrons
3 period

4 0

3 years ago

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.8

Lubov Fominskaja [6]

Complete Question

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added, you notice that the concentration of the unknown acid was 0.1 M. What is the pKa of your unknown acid?

Answer:

The pK_a value is $pK_a =7.82$

Explanation:

From the question we are told

The volume of base is $V_B = 26.mL = 0.0260L$

The pH of solution is $pH = 7.82$

The concentration of the acid is $C_A = 0.1M$

From the pH we can see that the titration is between a strong base and a weak acid

Let assume that the the volume of acid is $V_A = 18mL= 0.018L$

Generally the concentration of base

$C_B = \frac{C_AV_A}{C_B}$

Substituting value

$C_B = \frac{0.1 * 0.01800}{0.0260}$

$C_B= 0.0692M$

When 13mL of the base is added a buffer is formed

The chemical equation of the reaction is

$HA_{(aq)} + OH^-_{(aq)} --------> A^{+}_{(aq)} + H_2 O_{(l)}$

Now before the reaction the number of mole of base is

$No \ of \ moles[N_B] = C_B * V_B$

Substituting value

$N_B = 0.01300 * 0.0692$

$= 0.0009 \ moles$

Now before the reaction the number of mole of acid is

$No \ of \ moles = C_B * V_B$

Substituting value

$N_A = 0.01800 *0.1$

$= 0.001800 \ moles$

Now after the reaction the number of moles of base is zero i.e has been used up

this mathematically represented as

$N_B ' = N_B - N_B = 0$

The number of moles of acid is

$N_A ' = N_A - N_B$

$= 0.0009\ moles$

The pH of this reaction can be mathematically represented as

$pH = pK_a + log \frac{[base]}{[acid]}$

Substituting values

$7.82 = pK_a +log \frac{0.0009}{0.0009}$

$pK_a =7.82$

8 0

3 years ago

The general electron configuration for atoms of all elements in group 5a is

stiks02 [169]

Well the elements would be N, P, As, Sb, and Bi. Their electron configuration would be N= [He] 2s2 2p3, P= 1s2 2s2 2p6 3s2 3p3, As= [Ar] 3d10 4s2 4p3, Sb= Kr 4d10 5s2 5p3, and Bi= Xe 4f14 5d10 6s2 6p3.<span />

7 0

4 years ago

a student adds 3.5 moles of solute to enough water to make a 1500mL solution. what is the concentration?

aksik [14]

<h2>Hello!</h2>

The answer is:

$MolarConcentration=\frac{3.5moles}{volume(1.5L)}=2.33molar$

Since there is not information about the solute but only its mass, we need to assume that we are calculating the molar concentration of a solution or molarity. So, need to use the following formula:

$MolarConcentration=\frac{mass(solute)}{volume(solution)}$

Now, we know that the mass of the solute is equal 3.5 moles and the volume is equal to 1500 mL or 1.5L

Then, substituting into the equation, we have:

$MolarConcentration=\frac{3.5moles}{1.5L}=2.33molar$

Have a nice day!

7 0

3 years ago

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