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Stella [2.4K]
3 years ago
11

What would happen if less calcium chloride reacted and more unknown carbonate reacted?

Chemistry
1 answer:
Anit [1.1K]3 years ago
8 0
A white insoluble solid would appeaer
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On the graph, indicate the average kinetic energy of the population at T1 and T2.
kondaur [170]

Answer:

The average KE of the two graphs is where they intersect. Another name for temperature is kinetic energy because temperature is how fast atoms are moving and bumping into each other ( more bumping means higher temperature ).

6 0
3 years ago
What is the mass of 8.03 mole of NH3? (NH3 is 17g/mol)
lana [24]

Answer:

the mass of 8.03 mole of NH3 is 136.51 g

Explanation:

The computation of the mass is shown below:

As we know that

Mass = number of moles × molar mass

= 8.03 mol  × 17 g/mol

= 136.51 g

Hence, the mass of 8.03 mole of NH3 is 136.51 g

We simply multiplied the number of moles with the molar mass so that the mass could come

3 0
3 years ago
In a chemical reaction, a certain compound changes intoanother compound at a rate proportional to the unchanged amount. ifinitia
Nadya [2.5K]

No it will be a 10% of that balance so take 16 * 10% and then take that answer and divide bye 70%

7 0
4 years ago
Which nucleic acid is responsible for controlling the type of cell that is created? a. pla b. dna crna
avanturin [10]
It would be B. DNA it contains the genetic information for synthesizing specific proteins and other structures that characterize a cell to be a particular cell.
3 0
3 years ago
The pyrolysis of ethane proceeds with an activation energy of about 300 kJ/mol. How much faster is the decomposition at 625°C th
Alona [7]

Answer:

The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_2 = rate of reaction at T_2

K_1 = rate of reaction at T_1

Ea = activation energy of the reaction

R = gas constant = 8.314 J/K mol

E_a=300 kJ/mol=300,000 J/mol

T_2=625^oC=898.15 K,T_1=525^oC=798.15 K

\log (\frac{K_2}{K_1})=\frac{300,000 J/mol}{2.303\times 8.314 J/K mol}[\frac{1}{798.15 K}-\frac{1}{898.15 K}]

\log (\frac{K_2}{K_1})=2.185666

K_2=153.344\times K_1

The decomposition of ethane is 153.344 times much faster at 625°C than at 525°C.

4 0
3 years ago
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