Chemistry, asked by Nithin2134, 8 months ago

plz answer these

3. In the reaction

CuO (s) + H2 (g) → Cu(s) + H2O (g)

(a) Name the oxidized substance.

(b) Name the reduced substance.

(c) Name the oxidizing agent.


4. Give reasons :

(a) White silver chloride turns grey in sunlight.

(b) Brown coloured copper powder on heating in air turns into black coloured substance.


5. Compound ‘X’ decomposes to form compound ‘Y’ and CO2 gas. Compound Y is used in manufacturing of cement.

(a) Name the compounds ‘X’ and ‘Y’.

(b) Write the chemical equation for this reaction.


6. A metal salt MX when exposed to light splits upto to form metal M and gas X2. Metal M is used to make ornaments whereas gas X2 is used in making bleaching powder. The salt MX is used in black & white photography.

(a) Identify the metal M and gas X2

(b)Identify MX.

(c) Write down the chemical reaction when salt MX is exposed to sunlight.


7. A metal strip X is dipped in blue coloured salt solution YSO4. After some time a layer of metal ‘Y’ is formed on metal strip X. Metal X is used in galvanization whereas metal Y is used for making electric wires.

(a) What could be metal ‘X’ and ‘Y’ ?

(b) Name the metal salt YSO4.

Answers

Answered by rehmatsara
1

The probability that one bulb taken out random is non-defective is \dfrac{49}{50}

50

49

.

Step-by-step explanation:

Given as :

The number of bulb in the box = 600

The number of defective bulb in box = 12

Let The probability that one bulb taken out random is non-defective = p(E)

According to question

As out of 600 bulbs, 12 bulbs are defective

So, The number of non-defective bulb = 600 - 12 = 588

Now, out of 600 bulbs, 1 bulb is chosen in {C_{1}}^{600}C

1

600

And favourable case of chosen non-defective = {C_{1}}^{588}C

1

588

So, probability = \frac{{C_{1}}^{588}}{{C_{1}}^{600}}

C

1

600

C

1

588

Or, p(E) = \dfrac{588}{600}

600

588

Or, p(E) = \dfrac{49}{50}

50

49

Hence, The probability that one bulb taken out random is non-defective is \dfrac{49}{50}

50

49

. Answer

Answered by Cosmique
5

3.

a) substance oxidized = H₂  ( hydrogen )

b) substance reduced = Cu O   ( copper oxide )

c) oxidizing agent = Cu O  ( copper oxide)

:::::: ( i had already answered your (4) and ( 6) questions with your other asked questions ) :::::::

5)

a) compound X = Ca CO₃     ( calcium carbonate)

compound Y = Ca O      ( calcium oxide)

b)  chemical reaction is as below:

calcium carbonate ( heated)  →→→calcium oxide +  carbon dioxide

Ca CO₃ (Δ) →→→→→→→ Ca O + CO₂

7)

a) metal X = zinc (Zn)

metal Y = copper (Cu)

b)

metal salt YSO₄ is copper sulphate ( CuSO₄) .

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