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2 years ago

6

Which two energy sources were used the most in

1 answer:

Naily [24]2 years ago

4 0

Answer:

The 1st and 2nd ones on the top

Explanation:

Hope this helps:)

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Increasing the concentration of a reactant shifts the position of a chemical equilibrium towards formation of more products. wha

Natalija [7]

<h2>The required "option is e)".</h2>

Explanation:

The Rate of formation of products depends on the concentration of reactants or the forward reaction increases on increasing the rate of the concentration.
There is no effect on the equilibrium rate when the concentration of reactants and products is constant.
Forward reaction slows down when the reactant concentration is decreased.
On increasing the amount or concentration of reactant the chemical equilibrium shifts towards formation of more products.
Hence, the forward reaction speeds up.

4 0

2 years ago

Is the mold on bread made up of cells

Harman [31]

Answer:

Yes.

Explanation:

Mold is a living organism. All living organisms are made up of cells.

5 0

2 years ago

? Match the states of matter to their properties. Drag the items on the left to the correct location on the right. solids Indefi

Fantom [35]

Answer:

Solids: definite shape and definite volume (highest density)

Liquid: indefinite shape and definite volume (glide past each other)

Gas: indefinite shape and indefinite volume (lowest density)

Explanation:

look at the answer

8 0

2 years ago

Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given

tekilochka [14]

Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

$K_a$ = $1.34*10^{-5$

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝) $\alpha = \sqrt{\frac{K_a}{C} }$

$\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }$

$\alpha = \sqrt{4.9629*10^{-5}}$

$\alpha = 7.044*10^{-3$

percentage% (∝) = $7.044*10^{-3}*100$

= 0.704%

For Part B; where Concentration of B = $7.84*10^{-2$ M

$\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }$

$\alpha = \sqrt{1.709*10^{-4} }$

$\alpha = 0.0130\\$

percentage% (∝) = 0.0130 × 100%

= 1.30%

For Part C; where Concentration of C= $1.92*10^{-2} M$

$\alpha = \sqrt{\frac{1.34*10^{-5}}{1.97*10^{-2}} }$

$\alpha = \sqrt{6.802*10^{-4}}$

$\alpha =0.02608$

percentage% (∝) = 0.02608 × 100%

= 2.60%

7 0

3 years ago

Balance the following equations:

lesya [120]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>_____________________________________</h2>

(I) $N_{2} + 3H_{2} --> 2NH_{3}$

(II) $P_{4} + 5O_{2} --> 2P_{2} O_{5}$

(III) $2NaF + Br_{2} -->2NaBr + F_{2}$

(IV) $2ZnS + 3O_{2} --> 2ZnO + 2SO_2$

(V) $Pb(NO_3)_2 + 2NaCl --> 2NaNO_3 + PbCl_2$

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

7 0

2 years ago