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Troyanec [42]
3 years ago
15

Why is the sky black in space

Chemistry
1 answer:
Shkiper50 [21]3 years ago
3 0

Answer:Since there is virtually nothing in space to scatter or re-radiate the light to our eye, we see no part of the light and the sky appears to be black.

Explanation:

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Please help will mark brainliest!
Llana [10]

Answer:

1. desert

2. grasslands

3. rainforest

4. tundra

5. temperate deciduous forest

6. coniferous forest

Explanation:

educated guess, good luck

8 0
3 years ago
See the picture to answer the question
amid [387]
Above is a potential energy curve of a reaction. It depicts conversion of reactant to product via transition state.

When a catalyst is added to the reaction system, energy barrier of reaction decreases.

It must be noted that energy barrier reaction is defined as energy difference between  reactant and transition state.

In present case, energy of reactant is 200 kj, while that of transition state (in absence of catalyst) is 650 kj

Thus, energy barrier of reaction is 650 - 200 = 450 kj

<span>Hence, system must absorb 450 kj of energy for the reaction to start, if no catalyst was used</span>

4 0
3 years ago
CO(g) + 12 O2(g) → CO2(g)The combustion of carbon monoxide is represented by the equation above.(a) Determine the value of the s
devlian [24]

Answer : The standard enthalpy change for the combustion of CO(g) is, -283 kJ/mol

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 combustion of CO will be,

CO(g)+\frac{1}{2}O_2(g)\rightarrow CO_2(g)    \Delta H_{rxn}=?

The intermediate balanced chemical reaction will be,

(1) C(s)+\frac{1}{2}O_2(g)\rightarrow CO(g)     \Delta H_1=-110.5kJ/mol

(2) C(s)+O_2(g)\rightarrow CO_2(g)     \Delta H_2=-393.5kJ/mol

Now we are reversing reaction 1 and then adding both the equations, we get :

(1) CO(g)\rightarrow C(s)+\frac{1}{2}O_2(g)     \Delta H_1=110.5kJ/mol

(2) C(s)+O_2(g)\rightarrow CO_2(g)     \Delta H_2=-393.5kJ/mol

The expression for enthalpy change for the reaction will be,

\Delta H_{rxn}=\Delta H_1+\Delta H_2

\Delta H_{rxn}=(110.5)+(-393.5)

\Delta H_{rxn}=-283kJ/mol

Therefore, the standard enthalpy change for the combustion of CO(g) is, -283 kJ/mol

6 0
4 years ago
Please help I will give brainliest!
aleksklad [387]
I believe it’s the 3rd option
6 0
3 years ago
This group of fungi forms spores in a round structure on the end of a hyphae
r-ruslan [8.4K]

Answer:

Is zygote fungi an option??

3 0
3 years ago
Read 2 more answers
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