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alexandr1967 [171]
4 years ago
7

Can you please solve this. ..above question​

Chemistry
1 answer:
Dennis_Churaev [7]4 years ago
6 0

Answer:

(dont know if this is correct took this last year)

Explanation:

1 hydrogen atom plus 1 chlorine atom is equal to 2 hydrochloric acids due to the fact that there are 2 atoms in both substances and the equation has to stay constant throughout according to the law of conservation of mass

same goes for 2 sulfate atoms  and 3 oxygen atoms

in order to balance this this the sulfate atoms have to take into account that their are 2 sulfate atoms and 6 oxygen atoms  so you have to add more sulfates to make them both equal 6 then balance it with the oxygens

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3 years ago
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1 year ago
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A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after
Rus_ich [418]

Answer:

pH = 9.25

Explanation:

Step 1: Data given

Volume of a 0.40 M NH4Cl solution = 200.0 mL = 0.200 L

Molarity of NaOH = 0.80 M

Volume of NaOH = 50.0 mL

The Kb of ammonia is 1.76×10^-5

Step 2: The balanced equation

NH4Cl + NaOH → NH3 + NaCl + H2O

Step 3: Calculate moles

Moles NH4Cl  = molarity *¨volume

Moles NH4Cl = 0.40 M * 0.200 L = 0.08 moles

Moles NaOH = 0.80 M * 0.050 L = 0.04 moles

Step 4: Calculate limiting reactant

NaOH will completely be consumed. (0.04 moles). NH44Cl is in excess. There will react 0.04 moles .there will remain 0.04 moles.

Step 5: Calculate moles NH3

Moles NH3 = for 0.4 moles NH4Cl we'll have 0.4 moles NH3

Step 6: Calculate molarity

Molarity = moles / volume

Molarity = 0.4 moles = 0.250 L

Molarity = 1.6 M

Step 7: Calculate pOH

pOH = pKb + log ([B+]/[BOH])

pOH = 4.75 + log (1.6 / 1.6)

pOH = 4.75

Step 8: Calculate pH

pH = 14 - 4.75

pH = 9.25

3 0
3 years ago
A chemist has one solution that is 20% salt and a second solution that is 45% salt. How many liters of each should be used in or
Free_Kalibri [48]

Answer:

8 liters of the first salt solution and 12 liters of the second salt solution is needed to get 20L of a solution that is 30% salt

Explanation:

Let x represent the liters of the first salt solution

Let y represent the liters of the second salt solution

x + y= 20......equation 1

x= 20-y

The first solution is 20% salt and the second solution is 45% salt

20/100x + 45/100 y= 30/100 × 20

0.2x + 0.45y= 0.3×20

0.2x + 0.45y= 6............equation 2

Substitute 20-y for x in equation 2

0.2(20-y) + 0.45y= 6

4-0.2y+ 0.45y= 6

4 + 0.25y= 6

0.25y= 6-4

0.25y= 2

y= 2/0.25

y= 8

Substitute y for 8 in equation 1

x + y= 20

x + 8=20

x= 20-8

x= 12

Hence 8 liters of the first salt solution and 12 liters of the second salt solution is needed to get 20L of a solution that is 30% salt.

4 0
3 years ago
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