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Elodia [21]
3 years ago
12

Q2

Chemistry
1 answer:
sergey [27]3 years ago
5 0

Answer: 1.00 M

Explanation:

Moles of Solute/Liters of Solution = molarity. Then you would do 0.500 Moles/ 0.500 L and you get 1.00 M.

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Do states need to be written in a chemical equation
Reptile [31]
Yes. because equation balances the number of particles.
4 0
3 years ago
When solutions of magnesium chloride and potassium phosphate are combined write the formula for the precipitate formed
Zielflug [23.3K]
The products will be magnesium phosphate and potassium chloride. You then have to watch a solubility chart to see which one of these is not soluable. In this case it is magnesium phosphate.
6 0
3 years ago
Part 1. A chemist reacted 18.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equat
Alika [10]

Answer:

Part 1

The mass of the NaCl that reacted with F₂ at 290.K and 1.5 atm is approximately 132.6 gams

Part 2

The mass of NaCl that can react with the same volume of gas at STP is approximately 93.77 grams

Explanation:

Part 1

The volume of F₂ gas in the reaction, V = 18.0 liters

The ideal gas equation is P·V = n·R·T

∴ n = P·V/(R·T)

The pressure, P = 1.5 atm

The temperature, T = 290 K

The universal gas constant, R = 0.0820573 L·atm/(mol·K)

∴ n = 1.5×18/(0.0820573 × 290) ≈ 1.134615

The number of moles of F₂ in the reaction n ≈ 1.134615 moles

The chemical reaction is given as follows;

F₂ + 2NaCl → Cl₂ + 2NaF

1 mole of F₂ reacts with 2 moles of NaCl

Therefore;

1.134615 moles of F₂ reacted with 2 × 1.134615 moles ≈ 2.26923 moles of NaCl

1 mole of NaCl = The molar mass of NaCl, MM = 58.44 g/mol

The mass, of 2.26923 moles of NaCl, m = Number of moles × MM

∴ m ≈ 2.26923 moles × 58.44 g/mol ≈ 132.6 grams

The mass of the NaCl ≈ 132.6 gams

Part 2

The volume occupied by 1 mole of all gases at STP = 22.4 l/mole

Therefore, the number of moles of F₂ in 18.0 L of F₂ = 18.0 L/(22.4 L/mole) ≈ 0.804 moles

Therefore;

The number of moles of NaCl, in the reaction n = 2 × The number of moles of F₂ ≈ 2×0.804 moles = 1.608 moles

The number of moles of NaCl, in the reaction n ≈ 1.608 moles

The mass of NaCl in the reaction, m = n × MM

∴ m ≈ 1.608 moles × 58.44 g/mol ≈ 93.97 grams

The mass of NaCl that can react with the same volume of gas at STP ≈ 93.77 grams

8 0
3 years ago
λmax for the π → π* transition in ethylene is 170 nm. Is the HOMO-LUMO energy difference in ethylene greater than or less than t
-Dominant- [34]

Answer:

the HOMO-LUMO energy difference in ethylene is greater than that of cis,trans−1,3−cyclooctadiene

Explanation:

The λmax is the wavelength of maximum absorption. We could use it to calculate the HOMO-LUMO energy difference as follows:

For ethylene

E= hc/λ= 6.63×10^-34×3×10^8/170×10^-9= 1.17×10^-18J

For cis,trans−1,3−cyclooctadiene

E= hc/λ=6.63×10^-34×3×10^8/230×10^-9=8.6×10^-19J

Therefore, the HOMO-LUMO energy difference in ethylene is greater than that of cis,trans−1,3−cyclooctadiene

8 0
3 years ago
What is the mass of a sample of water that takes 2000 kJ of energy to boil into steam at 373 K. The latent heat of vaporization
zzz [600]

Answer:

\boxed{\text{889 g}}

Explanation:

The formula relating the mass m of a sample and the heat q to vaporize it is

q = mL, where L is the latent heat of vaporization.

\begin{array}{rcl}2000 \times 10^{3} \text{ J} & = & m \times \dfrac{2.25 \times 10^{6} \text{ J}}{\text{1 kg}}\\\\m & = & \dfrac{2000 \times \times 10^{3}\text{ kg}}{2.25 \times 10^{6}}\\ & = & \text{0.889 kg}\\\\ & = & \text{889 g}\\\end{array}\\\text{The mass of water is $\boxed{\textbf{889 g}}$}

5 0
3 years ago
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