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kipiarov [429]
3 years ago
5

An aqueous solution of potassium chloride is mixed with an aqueous solution of sodium nitrate.The complete ionic equation contai

ns which of the following species (when balanced in standard form)?
a. Cl-(aq)
b.KNO3(aq)
c.3NO-3(aq)
d.2Na+(aq)
e.2K+(aq)
Chemistry
1 answer:
Alexeev081 [22]3 years ago
4 0

Answer: Cl^-(aq) ions are present in the complete ionic equation

Explanation:

A double displacement reaction is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.

The balanced chemical reaction will be:

KCl(aq)+NaNO_3(aq)\rightarrow NaCl(aq)+KNO_3(aq)

The complete ionic equation can be written in terms of ions as:

K^+(aq)+Cl^-(aq)+Na^+(aq)+NO_3^-(aq)\rightarrow Na^+(aq)+Cl^(aq)+K^+(aq)+NO_3^-(aq)

Thus the complete ionic equation contains Cl^-(aq) species.

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2 years ago
A solution of ammonia has a pH of 11.8. What is the concentration of OH– ions in the solution?
balu736 [363]
PH + pOH = 14

11.8 + pOH = 14

pOH = 14 - 11.8

pOH = 2.2

[OH-] = 10 ^- pOH

[OH-] = 10 ^- 2.2

[OH-] = <span>6.33 x 10^-3 M
</span>
Answer B

hope this helps!





3 0
3 years ago
Read 2 more answers
Calculate the number of moles of NaOH contained in 250. mL of a 0.05M solution.
steposvetlana [31]

Answer:

0.0125mol

Explanation:

Molarity (M) = number of moles (n) ÷ volume (V)

n = Molarity × Volume

According to this question, a 0.05M solution contains 250 mL of NaOH. The volume in litres is as follows:

1000mL = 1L

250mL = 250/1000

= 0.250L

n = 0.05 × 0.250

n = 0.0125

The number of moles of NaOH is 0.0125mol.

3 0
2 years ago
Submit At 25.0 C, a 10.00 L vessel is filled with 5.00 atm of Gas A and 7.89 atm of Gas B. What is the mole fraction of Gas B?
Firlakuza [10]

Answer:

the mole fraction of Gas B is xB= 0.612 (61.2%)

Explanation:

Assuming ideal gas behaviour of A and B, then

pA*V=nA*R*T

pB*V=nB*R*T

where

V= volume = 10 L

T= temperature= 25°C= 298 K

pA and pB= partial pressures of A and B respectively = 5 atm and 7.89 atm

R= ideal gas constant = 0.082 atm*L/(mol*K)

therefore

nA= (pA*V)/(R*T) = 5 atm* 10 L /(0.082 atm*L/(mol*K) * 298 K) = 2.04 mole

nB= (pB*V)/(R*T) = 7.89 atm* 10 L /(0.082 atm*L/(mol*K) * 298 K) = 3.22 mole

therefore the total number of moles is

n = nA +nB= 2.04 mole +  3.22 mole = 5.26 mole

the mole fraction of Gas B is then

xB= nB/n= 3.22 mole/5.26 mole = 0.612

xB= 0.612

Note

another way to obtain it is through Dalton's law

P=pB*xB , P = pA+pB → xB = pB/(pA+pB) = 7.69 atm/( 5 atm + 7.89 atm) = 0.612

5 0
3 years ago
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