All categories

3 years ago

Pick the odd one out from the following.

Chemistry

1 answer:

Sophie [7]3 years ago

3 0

Answer:

Smoke

Explanation:

All of those listed are under the group of matter, except for smoke, making it the odd one one.

The rough definition of matter means that it occupies space, but smoke does not.

There are 3 main groups, Solid, Liquid and Gas.

Milk is classified under liquid.

Carbon Dioxide is classified under gas.

Butter is classified under solid.

Since you cannot classify smoke under any of the 3 categories, it is the odd one out.

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When the following equation is balanced with lowest whole-number coefficients, what is the coefficient for NO(g)? ___NH3(g) + __

inn [45]

Answer:

Coefficient in front of the $NO$ in the balanced equation - 4

Explanation:

The unbalanced reaction is shown below as:-

$NH_3+O_2\rightarrow NO+H_2O$

On the left hand side,

There are 1 nitrogen atom and 3 hydrogen atoms and 3 oxygen atoms

On the right hand side,

There are 1 nitrogen atom and 2 hydrogen atoms and 2 oxygen atoms

Thus,

Right side, $H_2O$ must be multiplied by 6 and $NO$ by 4

Left side, $O_2$ is multiplied by 5 and $NH_3$ by 4 so to balance the whole reaction.

Thus, the balanced reaction is:-

$4NH_3+5O_2\rightarrow 4NO+6H_2O$

<u>Coefficient in front of the $NO$ in the balanced equation - 4</u>

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3 years ago

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Vilka [71]

Answer: There are 20 grams

Explanation:

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3 years ago

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Explanation:

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4 years ago

___ Au₂S₃ + ___ H₂ → ___ Au + ___ H₂S

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Answer:

2Au₂S₃ + 6H₂ → 4Au + 6H₂S

Explanation:

Balancing:

2Au₂S₃ + 6H₂ → 4Au + 6H₂S

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3 years ago

A 3.00-L flask is filled with gaseous ammonia, NH3. The gas pressure measured at 27.0 ∘C is 2.55 atm . Assuming ideal gas behavi

Whitepunk [10]

Answer : The mass of ammonia present in the flask in three significant figures are, 5.28 grams.

Solution :

Using ideal gas equation,

$PV=nRT\\\\PV=\frac{w}{M}\times RT$

where,

n = number of moles of gas

w = mass of ammonia gas = ?

P = pressure of the ammonia gas = 2.55 atm

T = temperature of the ammonia gas = $27^oC=273+27=300K$

M = molar mass of ammonia gas = 17 g/mole

R = gas constant = 0.0821 L.atm/mole.K

V = volume of ammonia gas = 3.00 L

Now put all the given values in the above equation, we get the mass of ammonia gas.

$(2.55atm)\times (3.00L)=\frac{w}{17g/mole}\times (0.0821L.atm/mole.K)\times (300K)$

$w=5.28g$

Therefore, the mass of ammonia present in the flask in three significant figures are, 5.28 grams.

8 0

4 years ago