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strojnjashka [21]
3 years ago
9

C12H22O11 + 11 H2SO4 12 C + 11 H2SO4 + 11 H2O

Chemistry
1 answer:
Vladimir [108]3 years ago
4 0

Answer:

B: The sulfuric acid is not consumed or react with the reactant.

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The electron config for 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p3​
nadya68 [22]

Answer:

I don't understand what you are asking

3 0
3 years ago
How many molecules are in the quantities below.? 2.0 moles
matrenka [14]
Mole=number of molecules/6.02x10²³
mole=2
number of molecules= 2x6.02x10²³
number of molecules=12.04x10²³
5 0
3 years ago
Calculate the equilibrium constant of the reaction below if the pressures are 1.0atm, 2.0 atm, and 1.0 atm respectively. PCl3 +
Makovka662 [10]

Answer:

K = 0.5

Explanation:

Based on the reaction:

PCl₃ + Cl₂ ⇄ PCl₅

The equilibrium constant, K, is defined as:

K = P PCl₅ / P PCl₃ * P Cl₂

<em>Where P represent the pressure at the equilibrium for each one of the gases involved in the equilibrium.</em>

<em />

As:

P PCl₅ = 1.0atm

P PCl₃ = 1.0atm

P Cl₂ = 2.0atm

K = 1.0atm / 1.0atm * 2.0atm

<h3>K = 0.5</h3>
7 0
2 years ago
a steel cylinder is left out in the sun the pressure at 2:00 pm was 1014 kpa at 33.0 c at 7:00 am at a temperature of 21.4c what
e-lub [12.9K]
The easiest way is to use the Law of Gay-Lussac. This law states that there is a direct relation between the temperature in Kelvin of a gas and the pressure.

Then, namig p the pressure and T the temperature in Kelvin and using subscripts for every state:

p/T is constant ==> p_1 / T_1 = p_2/T_2

From which you obtain:

p_2 = [p_1 / T_1] * T_2

T_1 = 33.0 + 273.15 = 306.15 K
T _2 = 21.4 + 273.15 = 294.55 K

p_1 = 1014 kPa

p_2 = 1014 kPa * 294.55 K / 306.15 K = 975.6 kPa
5 0
3 years ago
What’s the formula for Lithium acetate
notsponge [240]

Answer:

The formula for lithium acetate is CH3COOLi

Explanation:

The formula for lithium acetate is obtained by replacing the hydrogen atom bonding to the oxygen atom in acetic acid with Li as shown below:

CH3COOH + LiOH —> CH3COOLi + H2O

5 0
3 years ago
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