Tag: let's play with water

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Questions Related to let's play with water

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

$14$ g of element X combines with $16$ g of oxygen. On the basis of this information, which of the following is a correct statement?

  1. The element X could have an atomic weight of $7$ amu and its oxide is $XO$
  2. The element X could have an atomic weight of $14$ amu and its oxide is $\displaystyle X _{2}O$
  3. The element X could have an atomic weight of $7$ amu and its oxide is $\displaystyle X _{2}O$
  4. The element X could have an atomic weight of $14$ amu and its oxide is $\displaystyle XO _{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$14$ g of an element X combines with $16$ g of oxygen, then the element X could have an atomic weight of $7$ amu and its oxide is $X _2O $.
When the atomic weight is $7$, $14$ g will corresponds to $2$ moles.
$16$ g of oxygen corresponds to $1$ mole. 

The formula $X _2O $ suggests the valency of $1$ for X.

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

The % loss in mass after heating a pure sample of potassium chlorate (Mol. mass = 122.5) will be:

  1. 12.25

  2. 24.50

  3. 39.17

  4. 49.0
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$2KClO _3\rightarrow 2KCl+3O _2$
$2\times 122.5$ grams shows a wieght loss of  $3\times32 grams$
So 245 grams of $KClO _3$ is 100%
96 grams of $O _2$ is X%
$x=\frac{100\times 96}{245}=39.17$.
Hence option C is correct.

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

The volume of oxygen at NTP evolved when 1.70 g of sodium nitrate is heated to a constant mass is: 

  1. 0.112 litre

  2. 0224 litre

  3. 22.4 litre

  4. 11.2 litre
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The chemical equation is:

$2NaNO _3(s)→  2NaNO _2 (s)+ O _2(g)$

2 moles of NaNO$ _3$ give one mole of Oxygen/22.4 L oxygen at STP

weight of NaNO$ _3$ = 85g

2 moles of NaNO$ _3$ weigh = 2(23+14+48) = 170g

170 g of NaNO$ _3$ will produce 22.4 L Oxygen at STP

​1.70 g will produce $= \cfrac{1.70}{170}×22.4 =0.224$ litres

Hence, the correct option is $\text{B}$
Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

The mass of residue left after strongly heating 1.38 g of silver carbonate will be: 

  1. 1.16 g

  2. 1.33 g

  3. 2.66 g

  4. 1.08 g
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$Ag _2CO _3\rightarrow Ag _2O+CO _2(g)$
I mole of $Ag _2CO _3$ (275.74) grams  on decomposition gives a mass residue of 231.73 grams of $Ag _2O$ remaning mass is liberated as $CO _2$ gas.
If 1.38 grams $Ag _2CO _3$ is hetaed it produces X grams of mass residue of  $Ag _2O$
$X=\frac{1.38\times 231.739}{275.745}=1.16$ grams
Hence option A is correct.

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

At NTP, 10 litre of hydrogen sulphide gas reacted with 10 litre of sulphur dioxide gas. The volume of gas, after the reaction is complete, would be:

  1. 5 litre

  2. 10 litre

  3. 15 litre

  4. 20 litre
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$2H _2S+SO2\rightarrow 2H _2O+3S$

According to the equation $2\times 22.4 $ litre of  $H _2S$ gas reacts with $22.4$ litre of $SO _2$
So $5$ litre of $H _2S$ reacts with $10$ litre of $SO _2$
Hence option $A$ is correct.

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

The process of obtaining salt from sea water involves:

  1. evaporation

  2. condensation

  3. sublimation

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The process of obtaining salt from sea water involves the process evaporation is used an a large scale to obtain common salt from sea water. Sea water is trapped in shallow lakes and allowed to stand there. The heat of sun gradually evaporates water in the shallow lakes and common salt is left behind as a solid. Sea water has many salts dissolved in it.
Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

Changing of liquid into vapours without heating is called:

  1. freezing

  2. steaming

  3. evaporation

  4. none of above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Changing of liquid into vapours is called evaporation and it causes cooling.
And in freezing the liquid changes into solid and causes cooling.
And in steaming the liquid boils and changes in vapour but causes heating
Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

Rainfall is related to which process?

  1. Evaporation

  2. Condensation

  3. Sublimation

  4. Heating
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

The water vapor rises in the atmosphere due to heating of water by sunlight and there it cools down and forms tiny water droplets through something called condensation. When they all combine together, they grow bigger and are too heavy to stay up there in the air. This is when they will fall to the ground as rain, or maybe snow or hail by gravity.

Sublimation is not involved in rainfall process

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

The thermal resistance of heat transfer is low in:

  1. Drop wise condensation

  2. Filmwise condensation

  3. Bulk wise condensation

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The thermal resistance of heat transfer is low in drop wise condensation.


 Thermal resistance is a heat property and a measurement of a temperature difference by which an object or material resists a heat flow.