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baherus [9]
3 years ago
14

I need the answer in 15 mins pls urjent

Chemistry
1 answer:
Sedbober [7]3 years ago
7 0

Answer:

I don't know I'm sorry I will tell you another answer asks me to

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Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left
Vera_Pavlovna [14]
So C12 is the limiting reactant and P4 is the excess
Mass P4 consumed= 0.31 mol X 123.9 g/mol =38.41 g P4 consumed.
Hope i helped
6 0
3 years ago
A TLC plate showed 2 spots with Rf values of 0.25 and 0.26. The plate was removed from the developing chamber, the residual solv
Katena32 [7]

Answer:

See explanation

Explanation:

TLC is a chromatographic method in which the solute is spotted on a plate and the plate is placed in an air tight chamber containing a solvent. The solvent is maintained below the level of the spot. The capillary movement of the solvent through the plate achieves the required separation.

If two spots have Rf values of 0.25 and 0.26 respectively and then the plate was removed from the developing chamber, subsequently, the residual solvent was allowed to evaporate from the plate, and then the plate was returned to the developing chamber.

It will be observed after the second development is complete that the new Rf values will be 0.50 and 0.52 respectively. It will just be as though the second chromatogram picked up from where the first chromatogram stopped.

4 0
3 years ago
15. What volume of CCI, (d = 1.6 g/cc) contain
anastassius [24]

Answer:

\boxed{\text{(3) 9.6 L}}

Explanation:

1. Moles of CCl₄

n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}

2. Molar mass of CCl₄

MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol

3. Mass of CCl₄

m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}

4. Volume of CCl₄

V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}

4 0
3 years ago
1. Related to the number of particles in a gram of
Romashka-Z-Leto [24]

Answer:

Avogadro's number or Avogardro’s constant

Explanation:

I’m pretty sure this is correct if it’s not I’m sorry lol.

6 0
2 years ago
At equilibrium the partial pressures of N2O4 and NO2 are 0.35 atm and 4.3 atm. What is the Kp
ale4655 [162]

Answer:

Kp = 52.83

Explanation:

6 0
3 years ago
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