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musickatia [10]
3 years ago
7

2NH4Cl(s)+Ba(OH)2⋅8H2O(s)→2NH3(aq)+BaCl2(aq)+10H2O(l) The ΔH for this reaction is 54.8 kJ . How much energy would be absorbed if

24.7 g of NH4Cl reacts?
Chemistry
1 answer:
LekaFEV [45]3 years ago
5 0
1) Chemical equation

<span>2NH4Cl(s)+Ba(OH)2⋅8H2O(s)→2NH3(aq)+BaCl2(aq)+10H2O(l)

2) Stoichiometric ratios

2 mol NH4Cl(s) : 54.8 KJ

3) Convert 24.7 g of NH4Cl into number of moles, using the molar mass

molar mass of NH4Cl = 14 g/mol + 4*1 g/mol + 35.5 g/mol = 53.5 g/mol

number of moles = mass in grams / molar mass

number of moles = 24.7 g / 53.5 g/mol = 0.462 moles

4) Use proportions:

2 moles NH4Cl / 54.8 kJ = 0.462 moles / x

=> x = 0.462 moles * 54.8 kJ / 2 moles = 12.7 kJ

Answer: 12.7 kJ</span>
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exis [7]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.

a) We need to find the major species from A to F.

Major Species at A:

1. Na_{2} CO_{3}

Major Species at B:

1. Na_{2} CO_{3}

2. NaHCO_{3}

Major Species at C:

1. NaHCO_{3}

Major Species at D:

1. NaHCO_{3}

2. H_{2}CO_{3}

Major Species at E:

1. H_{2}CO_{3}

Major Species at F:

1. H_{2}CO_{3}

b) pH calculation:

At Halfway point B:

pH = pKa_{1} + log[CO_{3}.^{-2}]/[HCO_{3}.^{-1}]

pH = pKa_{1} = 6.35

Similarly, at halfway point D.  

At point D,

pH = pKa_{2} + log [HCO_{3}.^{-1}]/[H2CO_{3}]

pH = pKa_{2} = 10.33

8 0
3 years ago
After 2.0s, Isabela . was riding her bicycle at 3.0m/s on a straight path. After 5.0s she was moving 5.4m/s. What was her accele
AysviL [449]

Answer:

Her acceleration was  0.8 m/s².

Explanation:

Given data:

Final speed v₂ = 5.4 m/s

Initial speed v₁= 3.0 m/s

Initial time t₁ = 2 s

Final time t₂ = 5 s

Acceleration = ?

Solution:

acceleration = (v₂- v₁) / (t₂ -  t₁)

acceleration = 5.4 - 3 / 5-2

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pashok25 [27]

Answer:

Asnwer to your question

Explanation:

Can you add a picture or anything like that?

5 0
3 years ago
Information related to a titration experiment is given in the balanced equation and table below.
Scorpion4ik [409]

Answer:

0.24M

Explanation:

The equation for the reaction is given below:

H2SO4 + 2KOH → K2SO4 + 2H2O

From the equation above, we obtained the following information:

nA (mole of acid) = 1

nB (mole of base) = 2

Data obtained from the question include:

Va (volume of the acid) = 12mL

Ca (concentration of the acid) =?

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The Ca (concentration of the acid) can be obtained as follow:

CaVa/CbVb = nA/nB

Ca x 12 / 0.16 x 36 = 1 /2

Cross multiply to express in linear form as shown below:

Ca x 12 x 2 = 0.16 x 36

Divide both side by 12 x 2

Ca = 0.16 x 36/ 12 x 2

Ca = 0.24M

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3 years ago
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