All categories

3 years ago

37 moles of manganese bromide how many molecules?

Chemistry

1 answer:

DerKrebs [107]3 years ago

8 0

214.746049. I converted between grams Manganese

You might be interested in

The temperature of a sample of liquid water changes from 50°C to 30°C. Which statement best explains the change

KiRa [710]

Answer:

The molecules move slower than the temp of 50°c and their average kinetic energy decreases.

Explanation:

kinetic energy cannot increase as temperature is reduced. molecules will still move with reduced motion.

7 0

2 years ago

What is the name if this compound? C6H5-C-H=O

brilliants [131]

Answer:

<u><em>Benzaldehyde </em></u>(C6H5CHO) is an organic compound consisting of a benzene ring with a formyl substituent. It is the simplest aromatic aldehyde and one of the most industrially useful.

8 0

2 years ago

The dissolution of barium hydroxide in water is an exothermic process. Which of the following statements is correct?A) The entha

madreJ [45]

Answer:

Explanation:

An exothermic reaction is one in which heat is liberated to the surrounding by a reaction.In this case,the enthalpy of the product is less than that of the reactant and such a reaction has a negative enthalpy (ΔH).

8 0

3 years ago

Read 2 more answers

A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$