All categories

3 years ago

Balanced equation of magnesium metal and hydrochloric acid

Chemistry

1 answer:

Setler79 [48]3 years ago

6 0

The balanced chemical equation is,

2Mg+2HCl→2MgCl+H2↑

You might be interested in

Which of the following is a common property of both strong acid and strong base?

Gekata [30.6K]

Answer:please can you help me with the last question that I asked I will help you if you help me please help

Explanation:

3 0

3 years ago

Read 2 more answers

9. Which of Earth’s sphere consists of all living things? A. hydrosphere B. geosphere C. biosphere

Anna007 [38]

Answer:

i think its b

Explanation:

i took it

3 0

3 years ago

How many moles are in 125g of water?

ki77a [65]

To calculate number of moles, all you do is divide the given mass by the molecular molar mass:

<span>i.e. 125g / 18g = 6.94444g </span>

<span>Therefore, your answer is (a) 6.94 g</span>

7 0

3 years ago

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔

Vilka [71]

Answer : The correct option is, (C) 1.1

Solution : Given,

Initial moles of $N_2O_4$ = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The given equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initially c 0

At equilibrium $(c-c\alpha)$ $2c\alpha$

The expression of $K_c$ will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

where,

$\alpha$ = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

$K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}$

$K_c=1.066\aprrox 1.1$

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0

3 years ago

a 10.99g sample of NaBr contains 22.34% Na by mass. Considering the law of constant composition (define proportions), how many g

leonid [27]

Given :

A 10.99 g sample of NaBr contains 22.34% Na by mass.

To Find :

How many grams of sodium does a 9.77g sample of sodium bromine contain.

Solution :

By law of constant composition , in any given chemical compound, the elements always combine in the same proportion with each other.

Therefore , percentage of Na by mass in NaBr will be same for every amount .

Percentage of Na in 9.77 g NaBr is 22.34 % too .

Gram of Na = $9.77\times \dfrac{22.34}{100}=2.18\ g$ .

Hence , this is the required solution .

7 0

3 years ago