What do you think he will found out when he swings the pendulum for 15 seconds

Molecular motion is a measure of the movement of the molecules in a substance. Molecular motion is substantially different betwe

Ainat [17]

The answer would be "air, wood". Gases have the highest Kinetic energy and least Potential energy. Liquids have the 2nd highest Kinetic energy and 2nd least potential energy and solids have the least kinetic energy and highest potential energy.

4 0

4 years ago

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Calculate the pH and pOH of 500.0 mL of a phosphate solution that is 0.285 M HPO42– and 0.285 M PO43–. (Ka for HPO42- = 4.2x10-1

jekas [21]

Answer: when concentrations of acid and base are same, pH = pKa

PH = 12.38 pOH = 1.62

Explanation: pKa= -log(Ka)= 12.38. PH + pOH = 14.00

8 0

3 years ago

A chemist makes of magnesium fluoride working solution by adding distilled water to of a stock solution of magnesium fluoride in

mojhsa [17]

The question is incomplete, here is the complete question:

A chemist makes 600. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a stock solution of 0.00154 mol/L magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.

Answer: The concentration of chemist's working solution is $5.90\times 10^{-4}M$

Explanation:

To calculate the molarity of the diluted solution (chemist's working solution), we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the stock magnesium fluoride solution

$M_2\text{ and }V_2$ are the molarity and volume of chemist's magnesium fluoride solution

We are given:

$M_1=0.00154M\\V_1=230mL\\M_2=?M\\V_2=600mL$

Putting values in above equation, we get:

$0.00154\times 230=M_2\times 600\\\\M_2=\frac{0.00154\times 230}{600}=5.90\times 10^{-4}M$

Hence, the concentration of chemist's working solution is $5.90\times 10^{-4}M$

8 0

3 years ago

CsH«N2(l) + CuCl(s)

bulgar [2K]

0.498 moles of copper(II) phthalocyanine would be produced by

the complete cyclotetramerization of 255 grams of phthalonitrile in the

presence of excess copper(ll) chloride.

Copper(ll) phthalocyanine (Cu(C₃₂H₁₆N₈)) is produced by the cyclotetramerization of phthalonitrile (C₈H₄N₂) according to the following reaction: 4 C₈H₄N₂(l) + CuCl₂(s) → Cu(C₃₂H₁₆N₈)(s) + Cl₂(g) How many moles of copper(II) phthalocyanine would be produced by the complete cyclotetramerization of 255 grams of phthalonitrile in the presence of excess copper(II) chloride?

Let's consider the following balanced equation.

4 C₈H₄N₂(l) + CuCl₂(s) → Cu(C₃₂H₁₆N₈)(s) + Cl₂(g)

The molar mass of C₈H₄N₂ is 128.13 g/mol. The moles corresponding to 255 g of C₈H₄N₂ are:

$255 g \times \frac{1mol}{128.13 g} = 1.99 mol$

The molar ratio of C₈H₄N₂ to Cu(C₃₂H₁₆N₈) is 4:1. The moles of Cu(C₃₂H₁₆N₈) produced from 1.99 moles of C₈H₄N₂ are:

$1.99 mol C_8H_4N_2 \times \frac{1molCu(C_{32}H_{16}N_8)}{4mol C_8H_4N_2} = 0.498 mol Cu(C_{32}H_{16}N_8)$

0.498 moles of copper(II) phthalocyanine would be produced by

the complete cyclotetramerization of 255 grams of phthalonitrile in the

presence of excess copper(ll) chloride.

You can learn more about stoichiometry here: brainly.com/question/22288091

8 0

2 years ago

A negative change in entropy indicates that...

vovangra [49]

Answer:

B. the products have a smaller number of available energy microstates than the reactants.

4 0

3 years ago

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