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Dmitriy789 [7]
3 years ago
8

How many molecules are in 5.00 grams of NH3

Chemistry
1 answer:
leva [86]3 years ago
8 0

Answer:

0.293590565643325

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Helppp!!!!!! pleaseeee
Ugo [173]

Answer:

A

Explanation:

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5 0
4 years ago
Given a fixed amount of gas held at constant pressure, calculate the volume it would occupy if a 2.00 L sample were cooled from
aliina [53]

Answer:

1.82 L

Explanation:

We are given the following information;

  • Initial volume as 2.0 L
  • Initial temperature as 60.0°C
  • New volume as 30.0 °C

We are required to determine the new volume;

From Charles's law;

\frac{V_1}{T_1}=\frac{V_2}{T_2}

Where, V_1 and V_2 are initial and new volume respectively, while T_1 and T_2 are initial and new temperatures respectively;

T_1= 333 K

T_2=303K

V_1 =2.0L

Rearranging the formula;

V_2=\frac{V_1T_2}{T_1}

    = \frac{(2.0L)(303K)}{333K} \\=1.820 L

Therefore, the new volume that would be occupied by the gas is 1.82 L

7 0
3 years ago
In another experiment, if 80 xo3 molecules react with 104 brz3 molecules how many br2 molecules will be produced which reactant
BaLLatris [955]

This is an incomplete question, here is a complete question.

The balanced chemical reaction is:

6XO_3+8BrZ_3\rightarrow 6XZ_4+4Br_2+9O_2

In another experiment, if 80 XO_3 molecules react with 104 BrZ_3 molecules. How many Br_2 molecules will be produced which reactant will be used up in the reaction.

Answer : The number of molecules of Br_2  will be, 52 molecules and BrZ_3 reactant will be used up in the reaction because it is a limiting reagent and it limits the formation of product.

Explanation :

The balanced chemical reaction is:

6XO_3+8BrZ_3\rightarrow 6XZ_4+4Br_2+9O_2

First we have to determine the limiting reagent.

From the balanced reaction we conclude that,

As, 8 molecules of BrZ_3 react with 6 molecule of XO_3

So, 104 molecules of BrZ_3 react with \frac{104}{8}\times 6=78 molecule of XO_3

From this we conclude that, XO_3 is an excess reagent because the given moles are greater than the required moles and BrZ_3 is a limiting reagent and it limits the formation of product.

Now we have to calculate the molecules of Br_2

From the reaction, we conclude that

As, 8 molecules of BrZ_3 react to give 4 molecules of Br_2

So, 104 molecules of BrZ_3 react to give \frac{104}{8}\times 4=52 molecules of Br_2

Hence, the number of molecules of Br_2  will be, 52 molecules and BrZ_3 reactant will be used up in the reaction because it is a limiting reagent and it limits the formation of product.

4 0
3 years ago
The 85.2-g sample of the compound x4o10 contains 48.0 g of oxygen atoms. what is the molar mass of element x?
Vedmedyk [2.9K]
X4O10
Let molar mass of X be y
molar mass =  4y + 10 x 16 = 4y+160

so, moles = 85.2 / (4y+160)

Moles of oxygen = 10 x [85.2 / (4y+160) ]
Mass of oxygen  = 16 x 10 x [85.2 / (4y+160) ]
which is 48.0

so, 48 = 16 x 10 x [85.2 / (4y+160) ]

Solve the equation to get y.

y = 31
5 0
3 years ago
I need help with this, please :00000
mojhsa [17]

theoretical yield of the reaction is 121.38 g of NH₃ (ammonia)

limiting reactant is N₂ (nitrogen)

excess reactant is H₂ (hydrogen)

Explanation:

We have the following chemical reaction:

N₂ + 3 H₂ → 2 NH₃

Now we calculate the number of moles of each reactant:

number of moles = mass / molar weight

number of moles of N₂ = 100 / 28 = 3.57 moles

number of moles of H₂ = 100 / 2 = 50 moles

From the chemical reaction we see that 3 moles of H₂ are reacting with 1 moles of N₂, so 50 moles of H₂ are reacting with 16.66 moles of N₂ but we only have 3.57 moles of  N₂ available, so the limiting reactant will be N₂ and the excess reactant will be H₂.

Knowing the chemical reaction and the limiting reactant we devise the following reasoning:

if          1 mole of N₂ produce 2 moles of NH₃

then    3.57 moles of N₂ produce X moles of NH₃

X = (3.57 × 2) / 1 = 7.14 moles of NH₃

mass = number of moles × molar weight

mass of NH₃ = 7.14 × 17 = 121.38 g

theoretical yield of the reaction is 121.38 g of NH₃

Learn more about:

limiting reactant

brainly.com/question/13979150

#learnwithBrainly

4 0
4 years ago
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