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3 years ago

Ionic bonding accurs when

Chemistry

1 answer:

Alja [10]3 years ago

3 0

Explanation:

An ionic bond is formed by the complete transfer of some electrons from one atom to another. The atom losing one or more electrons becomes a cation—a positively charged ion. The atom gaining one or more electron becomes an anion—a negatively charged ion.

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What is the concentration in molarity of S2O32- (aq) in a solution prepared by mixing 150 mLmL of 0.149 MM Na2S2O3 (aq) with eno

Artist 52 [7]

Answer:

0.0890 M

Explanation:

Since the concentration of KCl is irrelevant in this case, the concentration of Na2S2O3 can be determined using a simple dilution equation:

C1V1 = C2V2, where C1 = 0.149 M, V1 = 150 mL, V2 = 250 mL

C2 = 0.149 x 150/250

= 0.089 M

To determine the concentration of S2O32- (aq), consider the equation:

$Na_2S_2O_3 => 2Na^+_{(aq)} + S_2O^2^-_3_{(aq)}$

The concentration of Na2S2O3 and S2O32- (aq) is 1:1

Hence, the concentration in molarity of S2O32- (aq) is 0.089 M.

To 3 significant figures = 0.0890 M

7 0

3 years ago

What is a condensate?

Iteru [2.4K]

Answer:

A condensate is a liquid formed by condensation.

Explanation:

3 0

3 years ago

Read 2 more answers

Equivalent mass of AL2O3

Maru [420]

Answer:

$Mass = 102g$

Explanation:

Given

Compound: $Al_2O_3$

Required

Determine the equivalent mass

In the above compound, we have: 2 Al and 3 (O)

The atomic mass of Aluminium is:

$Al = 27$

The atomic mass of Oxygen is:

$(O) = 16$

So, the equivalent mass is:

$Mass = 2 * 27 + 3 * 16$

$Mass = 54 + 48$

$Mass = 102g$

5 0

3 years ago

A solution is made by mixing 50 ml of 2.0m k2hpo4 and 25 ml of 2.0m kh2po4. the solution is diluted to a final volume of 200 ml.

Volgvan

50 mL of 2.0 M of $K_2HPO_4$ and 25 mL of 2.0 M of $KH_2PO_4$ were mixed to make a solution

Final volume of the solution after dilution = 200 mL (given)

Final concentration of $K_2HPO_4$ , [ $K_2HPO_4$ ] = $\frac{50 mL\times 2 M}{200 mL} = 0.5 M$

Final concentration of $KH_2PO_4$ , [ $KH_2PO_4$ ] = $\frac{25 mL\times 2 M}{200 mL} = 0.25 M$

Using Hasselbach- Henderson equation:

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

$pka of KH_2PO_4 = 6.85$

Substituting the values:

$pH = 6.85+ log \frac{0.5}{0.25}$

$pH = 6.85+ log 2$

$pH = 6.85+ 0.3 = 7.15$

Hence, the $pH$ of the final solution is 7.15.

8 0

3 years ago

Read 2 more answers

Syringe initially holds a sample of gas with a volume of 285 mL at 355 K and 1.88 atm. To what temperature must the gas in the s

Akimi4 [234]

<h3>Answer:</h3>

720.535 K

<h3>Explanation:</h3>

Concept tested: Combined gas law

From the question we are given;

Initial volume, V1 of the gas is 285 mL
Initial pressure, P1 of the gas is 1.88 atm
Initial temperature, T1 of gas is 355 K
Final pressure of the gas, P2 = 2.50 atm
Final volume of the gas, V2 = 435 mL

We are required to calculate the new temperature, T2of the gas;

To solve the question we are going to use the combined gas law.
According to the combined gas law;

$\frac{P1V1}{T1}=\frac{P2V2}{T2}$

Rearranging the formula we can calculate the new temperature;

$T2=\frac{P2V2T1}{P1V1}$

Therefore;

$T2=\frac{(2.50atm)(435mL)(355K)}{(1.88atm)(285mL)}$

$T2=720.535K$

Therefore, the new temperature of the gas in the syringe must be 720.535 K

5 0

4 years ago