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3 years ago

15

The term meniscus refers to

2 answers:

valkas [14]3 years ago

8 0

Answer:

B

Explanation:

Is confirmed correct because i took the test and got a %100

Nastasia [14]3 years ago

4 0

Answer:

B) the curved surface of a liquid being measured

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When an aqueous solution of strontium chloride is added to an aqueous solution of potassium sulfate, a precipitation reaction oc

Georgia [21]

Answer:

$2Cl^{-}_{(aq)} + 2K^{-} _{(aq)} = 2KClx_{(aq)}$

Explanation:

The reaction is a precipitation reaction. In other words, the two aqueous solutions react to give the solid salt, potassium chloride (KCl) in this case.

$SrCl_{2} _{(aq)} + K_{2}SO_{4} _{(s)} = SrSO_{4} _{(aq)} + 2KCl_{(s)}$

Because the reaction product is a solid, the net ionic equation can be written as:

$2Cl^{-}_{(aq)} + 2K^{-} _{(aq)} = 2KClx_{(aq)}$

This is the resultant equation after removing spectator ions: sulfate and strontium.

3 0

3 years ago

Mazyrski [523]

Okay so your answer for n is

7 0

3 years ago

Ascorbic acid (vitamin

svlad2 [7]

Let us see the structure of ascorbic acid

As shown there is no COOH group however the OH group can lose a proton and forms conjugate base

The conjugate base formed is stabilized due to resonance

More the stability of conjugate base more the strength of acid

Hence ascorbic acid behaves as an acid

8 0

3 years ago

AlCl3 + Na NaCl + Al<br><br> Did Na change oxidation number?

Ilya [14]

$Al^{+III}Cl^{-I} _{3}+Na^{0}\rightarrow Na^{+I}Cl^{-I} + Al^{0}\\\\
Yes$

3 0

3 years ago

Be sure to answer all parts. What is the [H3O+] and the pH of a buffer that consists of 0.26 M HNO2 and 0.89 M KNO2? (K, of HNO2

Aleksandr-060686 [28]

Answer : The $H_3O^+$ ion concentration is, $1.12\times 10^{-3}M$ and the pH of a buffer is, 2.95

Explanation : Given,

$K_a=7.1\times 10^{-4}$

Concentration of $HNO_2$ (weak acid)= 0.26 M

Concentration of $KNO_2$ (conjugate base or salt)= 0.89 M

First we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (7.1\times 10^{-4})$

$pK_a=4-\log (7.1)$

$pK_a=3.15$

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[KNO_2]}{[HNO_2]}$

Now put all the given values in this expression, we get:

$pH=3.15+\log (\frac{0.89}{0.26})$

$pH=2.95$

The pH of a buffer is, 2.95

Now we have to calculate the $H_3O^+$ ion concentration.

$pH=-\log [H_3O^+]$

$2.95=-\log [H_3O^+]$

$[H_3O^+]=1.12\times 10^{-3}M$

The $H_3O^+$ ion concentration is, $1.12\times 10^{-3}M$

4 0

3 years ago