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3 years ago

What gas does photosynthesis produce

Chemistry

2 answers:

denis23 [38]3 years ago

8 0

The answer would be Oxygen. Hope this helps!

Serga [27]3 years ago

3 0

Oxygen is the answer

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Which is an example of a precipitation reaction?

netineya [11]

B. Precipitates are formed when two solutions combine to form another solution and a solid.

6 0

3 years ago

The elementary reaction 2H2O(g)↽−−⇀2H2(g)+O2(g) 2H2O(g)↽−−⇀2H2(g)+O2(g) proceeds at a certain temperature until the partial pres

Dima020 [189]

Answer:

$6.25\times 10^{-6}$ is the value of the equilibrium constant at this temperature.

Explanation:

Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of products to the partial pressures of reactants each raised to the power equal to their stoichiometric ratios. It is expressed as $K_{p}$

$2H_2O(g)\rightleftharpoons 2H_2(g)+O_2(g)$

Partial pressures at equilibrium:

$p^o_{H_2O}=0.070 atm$

$p^o_{H_2}=0.0035 atm$

$p^o_{O_2}=0.0025 atm$

The equilibrium constant in terms of pressures is given as:

$K_p=\frac{(p^o_{H_2})^2\times (p^o_{O_2})}{(p^o_{H_2O})62}$

$K_p=\frac{(0.0035 atm)^2\times 0.0025 atm}{(0.070 atm)^2}=6.25\times 10^{-6}$

$6.25\times 10^{-6}$ is the value of the equilibrium constant at this temperature.

5 0

3 years ago

Kung ang English ng Araw ay Sun ano naman ang Araw-Araw?

svlad2 [7]

Answer:

sun-sun char HAHAHAHAH EVERYDAY

4 0

2 years ago

Read 2 more answers

Calculate the standard emf for the following reaction:

krek1111 [17]

In order to solve this, we need to know the standard cell potentials of the half reaction from the given overall reaction.
The half reactions with their standard cell potentials are:
2ClO−3(aq) + 12H+(aq) + 10e- = Cl2(g) + 6H2O(l)
E = +1.47

Br(l) + 2e- = 2Br-
E = +1.065

We solve for the standard emf by subtracting the standard emf of the oxidation from the reducation, so:
1.47 - 1.065 = 0.405 V

4 0

3 years ago

For the reaction of hydrogen with iodine

fenix001 [56]

Answer:

$r_{H_2} = \frac{-1}{2} r_{HI}$

Explanation:

Hello!

In this case, considering the given chemical reaction:

$H_2(g) + I_2(g) \rightarrow 2HI(g)$

Thus, by applying the law of rate proportions, we can write:

$\frac{1}{-1} r_{H_2} = \frac{1}{-1}r_{i_2} = \frac{1}{2} r_{HI}$

Whereas the stoichiometric coefficients of reactants are negative due their disappearance and that of the product is positive due to its appearance. In such a way, when we relate the rate of disappearance of hydrogen gas to the rate of formation of hydrogen iodide, we obtain:

$r_{H_2} = \frac{-1}{2} r_{HI}$

Best regards!

8 0

2 years ago