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svlad2 [7]
3 years ago
6

The energy released in the formation of a compound from its elements is always _____ the energy required to decompose that compo

und into its elements. select one:
a. similar to
b. less than
c. identical to
d. greater than
Chemistry
1 answer:
maksim [4K]3 years ago
6 0
The answer is C.Identical to
You might be interested in
Copper metal (Cu) reacts with silver nitrate (AgNO3) in aqueous solution to form Ag and Cu(NO3)2. An excess of AgNO3 is present.
leonid [27]
Stoichiometry time! Remember to look at the equation for your molar ratios in other problems.

31.75 g Cu | 1 mol Cu | 2 mol Ag | 107.9 g Ag         6851.65
⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻   →   ⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻  = 107.9 g Ag
       ∅        | 63.5 g Cu | 1 mol Cu | 1 mol Ag              63.5


There's also a shorter way to do this: Notice the molar ratio from Cu to Ag, which is 1:2. When you plug in 31.75 into your molar mass for Cu, it equals 1/2 mol. That also means that you have 1 mol Ag because of the ratio, qhich you can then plug into your molar mass, getting 107.9 as well.
3 0
3 years ago
Read 2 more answers
On combustion of 0.2 gm of organic compound gave 0.147g of CO2, 0.12g of H2O and 74.6 ml of nitrogen gas at STP. Find emperical
zhuklara [117]

Answer:

0.2 is conpound Co2 STP.

Explanation:

on combustion of 0.2 gm of organic compound gave 0.147g of CO2, 0.12g of H2O and 74.6 ml of nitrogen gas at STP. Find emperical formula of compound.?

​

5 0
2 years ago
Name: _________________ Temperature o fwater_25_degreecent.YOU MUST SHOW ALL CALCULATIONS TO RECEIVE CREDIT FOR THEM! DATA ANALY
Anton [14]

Answer:

1. 0.02 M

2. 0.01 M

3. 4×10⁻⁶

Explanation:

We know that V₁S₁ = V₂S₂

1.

Concentration of HCl = 0.05 M

end point comes at = 10 ml

So, concentration of OH⁻(aq) = [OH⁻(aq)] ⇒ (0.05 × 10) ÷ 25 ⇒ 0.02 M

2.

2mol of OH⁻(aq) ≡ 1 mole of Ca²⁺(aq)

[Ca²⁺] = 0.02 ÷ 2 = 0.01 M

3.

K_{sp} = [Ca²⁺(aq)] [OH⁻(aq)]²

Ca(OH)₂ (aq) ⇄ Ca²⁺ (aq) + 2OH⁻ (aq)

K_{sp} = [0.01 × (0.02)²] = 4×10⁻⁶

4.

If reaction is exothermic which means heat energy will get evolved as a result temperature of the reaction media will get increased during the course of the reaction. If temperature is externally increased, the reaction will go backward to accumulate extra heat energy.

5.

K_{sp} value describes the solubility of a particular ionic compound. The higher the K_{sp} value, the higher the Solubility will be.

6.

This may be due to uncommon ion effect. The process of other ions (K⁺ or Na⁺) may increase the solubility

4 0
3 years ago
A 1.00 g sample of octane (C8H18) is burned in a bomb calorimeter with a heat capacity of 837J∘C that holds 1200. g of water at
lubasha [3.4K]

Answer:

The heat of combustion for 1.00 mol of octane is  -5485.7 kJ/mol

Explanation:

<u>Step 1:</u> Data given

Mass of octane = 1.00 grams

Heat capacity of calorimeter = 837 J/°C

Mass of water = 1200 grams

Temperature of water = 25.0°C

Final temperature : 33.2 °C

<u> Step 2:</u> Calculate heat absorbed by the calorimeter

q = c*ΔT

⇒ with c = the heat capacity of the calorimeter = 837 J/°C

⇒ with ΔT = The change of temperature = T2 - T1 = 33.2 - 25.0 : 8.2 °C

q = 837 * 8.2 = 6863.4 J

<u>Step 3:</u> Calculate heat absorbed by the water

q = m*c*ΔT

⇒ m = the mass of the water = 1200 grams

⇒ c = the specific heat of water = 4.184 J/g°C

⇒ ΔT = The change in temperature = T2 - T1 = 33.2 - 25  = 8.2 °C

q = 1200 * 4.184 * 8.2 =  41170.56 J

<u>Step 4</u>: Calculate the total heat

qcalorimeter + qwater = 6863.4 + 41170. 56 = 48033.96 J  = 48 kJ

Since this is an exothermic reaction, there is heat released. q is positive but ΔH is negative.

<u>Step 5</u>: Calculate moles of octane

Moles octane = 1.00 gram / 114.23 g/mol

Moles octane = 0.00875 moles

<u>Step 6:</u> Calculate heat combustion for 1.00 mol of octane

ΔH = -48 kJ / 0.00875 moles

ΔH = -5485.7 kJ/mol

The heat of combustion for 1.00 mol of octane is  -5485.7 kJ/mol

8 0
2 years ago
How many moles of nitrogen trifluoride (NF3) can be produced from 9.65 mole of Fluorine gas (F2)
user100 [1]

Answer:

6.43 moles of NF₃.

Explanation:

The balanced equation for the reaction is given below:

N₂ + 3F₂ —> 2NF₃

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of NF₃.

Finally, we shall determine the number of mole of nitrogen trifluoride (NF₃) produced by the reaction of 9.65 moles of Fluorine gas (F₂). This can be obtained as follow:

From the balanced equation above,

3 moles of F₂ reacted to produce 2 moles of NF₃.

Therefore, 9.65 moles of F₂ will react to to produce = (9.65 × 2)/3 = 6.43 moles of NF₃.

Thus, 6.43 moles of NF₃ were obtained from the reaction.

4 0
3 years ago
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