In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.

Given mass of sample compound = 630 g

Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;

56.7 % = 56.7/100 = 0.567

Mass of transition metal = 0.567 * 630 = 357.21 g

Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g

4 0

3 years ago

Why is it important to control the reaction rate of the chemical process

xxTIMURxx [149]

$Mg+2HCl=MgCl_2+H_{2(g)}$

The reaction creates dihydrogen, hence if it's uncontrolled it could lead to potentially dangerous amounts of gas being released at once.

8 0

3 years ago

Why is there day and night

Burka [1]

Because the moon orbits the earth and when the moon gets in front of the sun it blocks the sunlight and then it becomes night

6 0

3 years ago

nswer the following questions relating to HCl, CH3Cl, and CH3Br.HCl(g), can be prepared by the reaction of concentrated H2SO4(ag

alexdok [17]

Answer:

It is an example of double displacement reaction.

4.8 g of NaCl is needed to react.

Explanation:

Balanced reaction: $H_{2}SO_{4}(aq.)+2NaCl(s)\rightarrow 2HCl(g)+Na_{2}SO_{4}(aq.)$

Here, oxidation states of H, S, O, Na and Cl do not change during reaction. Hence it is not a redox reaction.

In this reaction, cations and anions of the reactants interchange their partners during reaction. Hence, it is an example of double displacement reaction.

As $H_{2}SO_{4}(aq.)$ remain in excess amount therefore NaCl (s) is the limiting reagent. Hence production of HCl entirely depends on amount of NaCl used.

Molar mass of HCl = 36.46 g/mol

So, 3.0 g of HCl = $\frac{3.0}{36.46}$ mol of HCl = 0.082 mol of HCl

According to balanced equation-

2 moles of HCl are produced from 2 moles of NaCl

So, 0.082 moles of HCl are produced from 0.082 moles of NaCl

Molar mass of NaCl = 58.44 g/mol

So, mass of 0.082 moles of NaCl = $(0.082\times 58.44)$ g = 4.8 g

Hence 4.8 g of NaCl is needed to react.

4 0

3 years ago