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2 years ago

Which states of matter can form solutions?

Chemistry

2 answers:

IrinaVladis [17]2 years ago

8 0

Answer:

Explanation:

spongebob cheese pants

Gala2k [10]2 years ago

4 0

Answer:

I believe D

Explanation:

im not super sure but I'm pretttyyy sure

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A doctor's Order is 0.125 g of ampicillin. The liquid suspension on hand contains 250 mg/s. How many milliliters (mL) of the sus

Molodets [167]

Answer:

c) 2.5 mL

Explanation:

Solution

Doctors order = 0.125g

and

The liquid suspension concentration = 250 mg/5ml

= 0.250g/5ml

Or 0.05g/ml

Amount of ml of suspension required = 0.125g/(0.05g/ml) = 2.5ml

7 0

2 years ago

Can you help me please

irina [24]

Answer:

Concentrated solutions

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2 years ago

Each chemical element has an atomic number and a mass number. the atomic number refers to the number of ___________________the e

creativ13 [48]

The atomic number tells us the number of protons and electrons.
The atomic mass tells us the weight of the nucleus, which is the amount of protons and neutrons

5 0

3 years ago

Select the compound that is most likely to increase the solubility of ZnSe when added to water.NaCnMgBr2NaClKClO4

vlada-n [284]

Answer:

NaCl.

Explanation:

In the solution, ZnSe ionizes to $Zn^2^+$ and $Se^2^-$ . Following reaction represents the ionization of ZnSe in solution -

$ZnSe$ ⇄ $Zn^2^+ + Se^2^-$

As we want to increase the solubility of ZnSe, we must decrease the concentration of dissociated ions so that the reaction continues to forward direction.

If we add NaCl to this solution, then we have $Na^+$ and $Cl^-$ in the solution which will be formed by the ionization of NaCl.

Now, $Zn^2^+$ in the solution will react with two $Cl^-$ ions to form $ZnCl_2$ as follows -

$Zn^2^++2Cl^-$ ⇄ $ZnCl_2$

Due to this reaction the concentration of $Zn^2^+$ will decrease in the solution and more ZnSe can be soluble in the solution.

6 0

2 years ago

The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. A

liq [111]

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

$p^{OH}$ =14-56.17

=8.823

The volume of the $NH_{3}$ = 40.00 ml

convert into the liter= 0.040L

The value of the concentrated $NH_{3}$ =0.10 M

The volume of the $NH_{4}Cl$ = 50.00 ml

convert into the liter= 0.050L

The value of concentrated $NH_{4}Cl$ = 0.10 M

The volume of the $H_{2}So_{4}$ = 30 ml

convert into the liter= 0.030L

The value of concentrated $H_2So_4$ =0.05 M

Calculating total volume=(0.40+0.050+0.030)

=0.120 L

calculating the new concentrated value of $NH_3$ = $\frac{0.10\times 0.040}{0.120}= 0.33 \ M$

calculating the new concentrated value of $NH_4Cl$ = $\frac{0.050\times 0.10}{0.120}= 0.04166 \ M$ calculating the new concentrated value of $H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M$ when 1 mol $H_2So_4$ produced 2 mols $H^{+}$ so, 0.0125 in $H_2So_4$ produced:

$=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}$

create the ICE table:

$NH_3 \ \ \ \ \ \ \ \ + H^{+} \ \ \ \ \ \ \longrightarrow NH_4^{+}$

I (m) 0.033(m) 0.025 0.04166

C -0.025 -0.025 + 0.025

E 8.3\times 10^{-3} 0 0.0667

now calculating pH:

when ph= 8.83:

$P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769$

5 0

2 years ago