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Tpy6a [65]
3 years ago
14

Pls help me!!!!!!!! Answer in order please.

Chemistry
1 answer:
larisa86 [58]3 years ago
8 0
Q. No. 1:
                a)  Endothermic

                b)  Exothermic

                c)  Endothermic

                d)  Exothermic

Q. No. 2:
                a)  Exothermic

                b)  Because heat is on product side (evolved to surrounding).

                c)  
                  
                          1400 Kj  =  was produced by 1 mole of C₂H₄
Then,
                          2100 Kj  =  will be produced by X moles of C₂H₄

Solving for X,
                          X  =  (2100 Kj × 1 mol)  ÷ 1400 Kj

                          X  =  1.5 moles of C₂H₄

                d)

                             
3 moles of O₂ produced  =  1400 Kj of Heat
Then,
                      1.5 moles of O₂ will produce  =  X Kj of Heat

Solving for X,
                                        X  = (1.5 mol × 1400 Kj) ÷3 mol

                                        X  =  700 Kj of Heat
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Can someone please help me and ill mark u as brainlest
kodGreya [7K]

Answer:

Sure i'll help u! :D

Explanation:

for #1 you do A. I'm not so sure on it tho

For #2 it should be C.

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What is Alfunim second element ? Please help !!!
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3 years ago
A sample of air at 58.6 kPa is expanded from 250 mL to 655 mL. If thetemperature remains constant, what is the final pressure in
kati45 [8]

Answer:

<h2>22.366 kPa</h2>

Explanation:

The final pressure can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the final pressure

P_2 =  \frac{P_1V_1}{V_2}  \\

From the question

58.6 kPa = 58600 Pa

We have

P_2 =  \frac{58600 \times 250}{655}  =  \frac{14650000}{655}  \\  = 22366.4122... \\  = 22366

We have the final answer as

<h3>22.366 kPa</h3>

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3 years ago
How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation? 2HBr(aq)+
Mkey [24]

Full Question:

A flask containing 420 Ml of 0.450 M HBr was accidentally knocked to the floor.?

How many grams of K2CO3 would you need to put on the spill to neutralize the acid according to the following equation?

2HBr(aq)+K2CO3(aq) ---> 2KBr(aq) + CO1(g) + H2O(l)

Answer:

13.1 g K2CO3 required to neutralize spill

Explanation:

2HBr(aq) + K2CO3(aq) → 2KBr(aq) + CO2(g) + H2O(l)

Number of moles = Volume * Molar Concentration

moles HBr= 0.42L x .45 M= 0.189 moles HBr

From the stoichiometry of the reaction;

1 mole of K2CO3 reacts  with 2 moles of HBr

1 mole = 2 mole

x mole = 0.189

x = 0.189 / 2 = 0.0945 moles

Mass = Number of moles * Molar mass

Mass = 0.0945 * 138.21  = 13.1 g

3 0
3 years ago
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