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MAXImum [283]
3 years ago
5

Don't gotta show proof but pls explain how so i know you not guessing!!!!!! will give brainliest

Chemistry
1 answer:
Feliz [49]3 years ago
5 0

Answer:

Other: How does the temperature at which plants are grown affect plant growth?

Explanation:

The only variable in the table that isn't a constant is the temperature at which the plants were grown. This means that the scientists planted the groups of seedlings in rooms with different temperatures. The scientists then recorded the amount of new leaves the plants had per week to see if the temperature of the room had an effect on plant growth.

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Another metal phosphate is cobalt phosphate. It will behave similar to calcium phosphate in an acid solution. What is the net io
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We are given with
Cobalt phosphate - CoPO4

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Density is an example of a​
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hope this helps if it do mark me as brainlest thank you !!
7 0
3 years ago
Will acid in your stomach will make you unhealthy. do you agree or disagree? explain.
Alex787 [66]
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7 0
3 years ago
How to draw Hess' Cycle for this question ?
NISA [10]

Answer : The standard enthalpy of formation of ethylene is, 51.8 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of C_2H_4 will be,

2C(s)+2H_2(g)\rightarrow C_2H_4(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)     \Delta H_1=-1411kJ/mole

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.7kJ/mole

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.9kJ/mole

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equations, we get :

(1) 2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)     \Delta H_1=+1411kJ/mole

(2) 2C(s)+2O_2(g)\rightarrow 2CO_2(g)    \Delta H_2=2\times (-393.7kJ/mole)=-787.4kJ/mole

(3) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=2\times (-285.9kJ/mole)=-571.8kJ/mole

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(+1411kJ/mole)+(-787.4kJ/mole)+(-571.8kJ/mole)

\Delta H=51.8kJ/mole

Therefore, the standard enthalpy of formation of ethylene is, 51.8 kJ/mole

7 0
3 years ago
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