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antoniya [11.8K]

3 years ago

15

In the reaction _S +3O2 → 2SO3, what coefficient should be placed in front of the S to balance the reaction? i need the answer

1 answer:

Y_Kistochka [10]3 years ago

7 0

2s + 3O2 -> 2SO3 meaning the answer is 2

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calculate the volume od a CO2 cartridge that has a pressure of 850 PSI at a temperature of 21ºc the cartridge contains 0.273 mol

gladu [14]

Answer is: volume of CO₂ is 0,113 dm³.
Ideal gas law = pV = nRT.
p = 850 PSI = 5860543,6992 Pa.
Psi <span>is the abbreviation of pound per square inch.
T = 21</span>°C = 294,15 K.
n = 0,273 mol.
R = 8,314 J/K·mol.
V = nRT ÷ p
V = 0,273 mol · 8,314 J/K·mol · 294,15 K ÷ 5860543,6992 Pa.
V = 0,00011 m³ = 0,113 dm³.

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3 years ago

Is slime a solid or liquid?

Ivan

Answer:

i think its a liquid..... im not that sure

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3 years ago

A magnesium (Mg) atom gives one electron to two atoms of another element, then it takes on a 2+ charge. The two atoms of the oth

Setler [38]

Answer:

See below

Explanation:

You missed the following elements, but any element with 7 valence electrons could behave like that: Cl, Br, I, F, etc

3 0

3 years ago

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Which mix will be homogeneous?

DIA [1.3K]

Answer:

B

Explanation:

food coloring and water are inseparable after mixed.

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At equilibrium, the concentrations of the products and reactants for the reaction, H2 (g) + I2 (g)  2 HI (g), are [H2] = 0.106

lana [24]

Answer:

The new equilibrium concentration of HI: <u>[HI] = 3.589 M</u>

Explanation:

Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M

Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M

<em>Given chemical reaction:</em> H₂(g) + I₂(g) → 2 HI(g)

The equilibrium constant ( $K_{c}$ ) for the given chemical reaction, is given by the equation:

$K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}$

<u><em>At the original equilibrium state:</em></u>

$K_{c} = \frac {(1.29\: M)^{2}}{(0.106\: M) \times (0.022\: M)}$

$K_{c} = \frac {1.6641}{0.002332} = 713.59$

<u><em>Therefore, at the new equilibrium state:</em></u>

$K_{c} = \frac {[HI]^{2}}{(0.95\: M) \times (0.019\: M)}$

$\Rightarrow K_{c} = 713.59 = \frac {[HI]^{2}}{0.01805}$

$\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88$

$\Rightarrow [HI] = \sqrt {12.88} = 3.589 M$

<u>Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M</u>

6 0

3 years ago