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3 years ago

15

What one of these reactions are a combustion a. Burning wood b. Decomposition of water c. Production of plants with the suns ene

rgy d. Rotting fruits and vegetables

2 answers:

sineoko [7]3 years ago

6 0

Combustion involves with heat. The answer is burning wood.

tatiyna3 years ago

4 0

C.Production of plants with the suns energy.

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Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =

MAVERICK [17]

<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{c}$

For a general chemical reaction:

$aA+bB\rightleftharpoons cC+dD$

The expression for $K_{eq}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

$2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)$

The expression of $K_c$ for above equation is:

$K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}$

We are given:

$[H_2S]_{eq}=0.671M$

$[O_2]_{eq}=0.587M$

$K_c=1.35$

Putting values in above expression, we get:

$1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}$

$[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M$

Hence, the equilibrium concentration of water is 0.597 M

8 0

3 years ago

1.26 * 10^4 + 2.50 * 10^4 in sceintific notation

eduard

Hello! I can help you with this. First, convert them into it’s written out standard form. 10^4 is 10,000. 10,00 * 1.26 is 12,600. 10,000 * 2.5 is 25,000. 12,600 + 25,000 = 37,600 or 3.76 * 10^4 in scientific notation. The answer in scientific notation is 3.76 * 10^4.

8 0

3 years ago

Cho 4g CuO vào dung dịch axit clohidric 10% thì phản ứng vừa đủ.

Sholpan [36]

Answer:

Explanation:

a. CuO+ 2HCl⇒CuCl2+ H2O

b. $n_{CuO}$ = $\frac{4}{80}$ = 0,05 (mol)

⇒ $n_{CuCl2}$ = $n_{CuO}$ =0,05 mol

⇒ $m_{CuCl2}$ = 0,05×135=6,75 (g)

c. $n_{HCl}$ =2× $n_{CuO}$ =0,1 (mol)

⇒ $m_{HCl}$ = 0,1×36,5= 3,65 (g)

⇒ $m_{dd HCl}$ = $\frac{m_{HCl}}{10}$ ×100=36,5 (g)

⇒ Nồng độ phần trăm dd sau phản ứng= Nồng độ % dd CuCl2= $\frac{m_{CuCl2} }{m_{dd HCl+ m_{CuO} } }$ ×100= $\frac{6,75}{36,5+4}$ ×100≈ 16,67%

8 0

3 years ago

To use the right hand rule, you should point your thumb in the same

Flauer [41]

The Correct Answer is Eletric Current I think

5 0

3 years ago

If 71.5 moles of an ideal gas is at 5.03 atm at 6.80 °C, what is the volume of the gas?

meriva

Use the clapeyron equation:

T in kelvin : 6.80 + 273 => 279.8 K

R = 0.082

n = 71.5 moles

P = 5.03 atm

Therefore:

P x V = n x R x T

5.03 x V = 71.5 x 0.082 x 279.8

5.03 x V = 1640.4674

V = 1640.4674 / 5.03

V = 326.13 L

hope ths helps!

5 0

3 years ago