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nata0808 [166]
3 years ago
12

What is the name of the gas produced when nitric acid is added to copper metal? please help! much appreciated!

Chemistry
1 answer:
Lemur [1.5K]3 years ago
8 0
The answer to your question is nitrogen dioxide

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How much heat is released as the temperature of 25.2 grams of iron is decreased from 72.1°C to 9.8°C? The specific heat of iron
prisoha [69]

Answer:

Q=-697.06\ J

Negative sign says that release of heat.

Explanation:

The expression for the calculation of the heat released or absorbed of a process is shown below as:-

Q=m\times C\times \Delta T

Where,  

\Delta H  is the heat released or absorbed

m is the mass

C is the specific heat capacity

\Delta T  is the temperature change

Thus, given that:-

Mass = 25.2 g

Specific heat = 0.444 J/g°C

\Delta T=9.8-72.1\ ^0C=-62.3\ ^0C

So,  

Q=25.2\times 0.444\times -62.3\ J=-697.06\ J

Negative sign says that release of heat.

8 0
3 years ago
Calculate the number of grams of carbon found in a 5 mole sample of carbon.
Ksenya-84 [330]

Answer:

60g

Explanation:

1mol of carbon has 12g

3 0
3 years ago
What is the volume of 4.00 moles of CO2 gas at STP?
likoan [24]

Answer:

89.6L

Explanation:

1mole of any gas occupies 22.4L. This simply means that,

1mole of CO2 occupies 22.4L at stp.

Therefore, 4moles of CO2 will occupy = 4 x 22.4 = 89.6L

8 0
3 years ago
In the compound cah2 calcium has an oxidation number of 2+ and hydrogen has an oxidation number of
sesenic [268]

The oxidation number of H is -1.

Sum of the oxidation numbers in each element = charge of the complex

CaH₂ has 1 Ca atom and 2H atoms. The charge of the complex is zero. Let’s say Oxidation number of H is "a".

Then,

<span>    (+2) + 2 x a = 0 </span>

<span>        +2  + 2a  = 0</span>

                  2a = -2

                    a = -1

Hence, the oxidation number of Hydrogen atom in CaH₂ is -1


7 0
3 years ago
Consider the second-order reaction:
kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after 4.53\times 10^{10} s is 0.00345 mol/L.

Explanation:

2HI(g)\rightarrow H_2(g)+I_2(g)&#10;

Rate Law: k[HI]^2&#10;

Rate constant of the reaction = k = 6.4\times 10^{-9} L/mol s

Order of the reaction = 2

Initial rate of reaction = R=1.6\times 10^{-7} Mol/L s

Initial concentration of HI =[A_o]

1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2

[A_o]=5 mol/L

Final concentration of HI after t = [A]

t = 4.53\times 10^{10} s

Integrated rate law for second order kinetics is given by:

\frac{1}{[A]}=kt+\frac{1}{[A_o]}

\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}

[A]=0.00345 mol/L

The concentration of HI after 4.53\times 10^{10} s is 0.00345 mol/L.

5 0
4 years ago
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