All categories

2 years ago

Classify each of the following reactions when only the reactants are given and fill in the products.

Chemistry

1 answer:

Nadya [2.5K]2 years ago

3 0

Answer:

CO2 + MgO

Explanation:

We want the number of each element on the Reactants (Left side) to be equal to the number of each element on the Product side (Right side).

You might be interested in

I need help with this question please someone tell me the answer for this

Andrej [43]

Aueyeowosy yo this is my answer

5 0

2 years ago

The worst feeling is when you are hurting so badly inside but not a single tear comes out.

guapka [62]

Answer:

chemistry wow

Explanation:

8 0

2 years ago

2Ag + H2S ? Ag2S + H2 Which statement is true about this chemical equation? A) Ag (silver) is oxidized and H (hydrogen) is reduc

nadezda [96]

Answer:

It is A).

Explanation:

Silver (Ag) goes from the pure metal to Ag+ losing 1 electron so it is oxidised.

The hydrogen ion gains electrons and is reduced.

6 0

2 years ago

Read 2 more answers

How many grams of lead(II) sulfate (303 g/mol) are needed to react with sodium chromate (162 g/mol) in order to produce 0.162 kg

Afina-wow [57]

Answer : The mass of $PbSO_4$ needed are, 1.515 grams.

Explanation :

First we have to calculate the mole of $PbCrO_4$ .

$\text{Moles of }PbCrO_4=\frac{\text{Mass of }PbCrO_4}{\text{Molar mass of }PbCrO_4}=\frac{0.162g}{323g/mole}=0.005mole$

Now we have to calculate the moles of $PbSO_4$ .

The balanced chemical reaction will be,

$PbSO_4+Na_2CrO_4\rightarrow PbCrO_4+Na_2SO_4[tex]From the balanced chemical reaction, we conclude thatAs, 1 mole of [tex]PbCrO_4$ produced from 1 mole of $PbSO_4$

So, 0.005 mole of $PbCrO_4$ produced from 0.005 mole of $PbSO_4$

Now we have to calculate the mass of $PbSO_4$

$\text{Mass of }PbSO_4=\text{Moles of }PbSO_4\times \text{Molar mass of }PbSO_4$

$\text{Mass of }PbSO_4=0.005mole\times 303g/mole=1.515g$

Therefore, the mass of $PbSO_4$ needed are, 1.515 grams.

6 0

2 years ago

The percent of remaining parent isotope in a radioactive decay process is 40 percent. How many half-lives have elapsed since the

muminat

Answer: Between 1 and 2.

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

$a=\frac{a_o}{2^n}$ ............(1)

where,

a = amount of reactant left after n-half lives = 40

$a_o$ = Initial amount of the reactant = 100

n = number of half lives

Putting in the values we get:

$40=\frac{100}{2^n}$

$2^n=2.5$

taking log on both sides

$nlog(2)=log(2.5)$

$n=1.32$

Thus half-lives that have elapsed is between 1 and 2

7 0

3 years ago