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3 years ago

Plz help!!!

1 answer:

7 0

a) The change here is that metallic iron is converted into ions and copper is deposited. This is called a displacement reaction.

b) $\text{Fe}$ is oxidised in this reaction.

c)$ \mathrm{Fe_{(s)}+ CuSO_{4(aq)} \rightarrow FeSO_{4(aq)} + Cu_{(s)}}$

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The following balanced equation shows the decomposition of water (H2O). According to the equation, how many oxygen atoms are in

EleoNora [17]

2 atoms of oxygen are reactants.

6 0

2 years ago

Read 2 more answers

How many ml of 0.5M of HNO3 would be needed to react with 85ml of 0.75M of KOH

olganol [36]

Answer: 127.5ml

Explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HNO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is KOH.

We are given:

$n_1=1\\M_1=0.5M\\V_1=?mL\\n_2=1\\M_2=0.75M\\V_2=85mL$

Putting values in above equation, we get:

$1\times 0.5\times V_1=1\times 0.75\times 85\\\\V_1=127.5mL$

Thus 127.5 ml of 0.5M of HNO3 would be needed to react with 85ml of 0.75M of KOH

6 0

2 years ago

How many moles of solute particles are present in 1 ml of aqueous 0.020 m (nh4)2co3?

Dima020 [189]

Vs = 1.0 mL = 0.001 L
c((NH4)2CO3) = <span>0.02 M
n(</span>(NH4)2CO3) = ?

For the purpose, here we will use the next equation:

c=n/V ⇒ n=cxV

n((NH4)2CO3) = 0.02M x 0.001L

n((NH4)2CO3) = 2x10⁻⁵ mole of (NH4)2CO3 is presented in the solution

6 0

3 years ago

10 POINTS.

enot [183]

There are a total of 40 valence electrons in the PCl5 Lewis structure. Remember when you draw the Lewis structure for PCl5 that Phosphorous (P) is in Period 3 on the Periodic table. This means that it can hold more than 8 valence electrons.

I hope this helps and please don't report

7 0

2 years ago

A chemist requires 0.881 mol Na2CO3 for a reaction. How many grams does this correspond to?

irga5000 [103]

Molar mass Na2CO3 = 106 g/mole
Using dimensional analysis:
0.787 moles Na2CO3 x 106 g/mole Na2CO3 = g Na2CO3
Answer = 83.4 g Na2CO3

5 0

2 years ago