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anastassius [24]
3 years ago
7

A piece of metal ore has a density of 1.72 g/mL. When a student places it into a graduated cylinder containing water, the liquid

level rises from 21.25 mL to 26.47 mL. Calculate the mass of the metal ore in grams and round the final answer. _______ g
Chemistry
1 answer:
Hunter-Best [27]3 years ago
5 0

Answer:

9g

Explanation:

Since..., Density = mass/volume

;Firstly calculate the volume change of the water when the metal is placed in,

;Volume = 26.47 - 21.25

;Volume = 5.22 mL

Hence the mass of the ore;

; Making mass the subject of formula in the above equation,

Mass = Density × Volume

; Mass = 1.72 × 5.22

; Hence the mass = 8.9784

= 9g

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Nataly_w [17]

Answer:

The answer to your question is a) N₂     b) 3.04 g of NH₃

Explanation:

Data

mass of H₂ = 2.5 g

mass of N₂ = 2.5 g

molar mass H₂ = 2.02 g

molar mass of N₂ = 28.02 g

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Balanced chemical reaction

                3H₂  +  1 N₂   ⇒   2NH₃

A)

Calculate the theoretical yield 3H₂ / N₂ = 3(2.02) / 28.02 = 0.22

Calculate the experimental yield H₂/N₂ = 2.5/2.5 = 1

Conclusion

The limiting reactant is N₂ (nitrogen) because the experimental proportion was higher than the theoretical proportion.

B)

             28.02 g of N₂ -------------------- (2 x 17.04) g of NH₃

               2.5 g of N₂    --------------------   x

                            x = (2.5 x 2 x 17.04) / 28.02

                            x = 85.2 / 28.02

                           x = 3.04 g of NH₃

6 0
3 years ago
Use the ideal-gas law to estimate the number of air molecules in your physics lab room, assuming all the air is N2. Assume a roo
stiv31 [10]

Answer:

4.13×10²⁷ molecules of N₂ are in the room

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Let's determine the volume of the room:

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6869.4 moles of N₂ are in the room

If we want to find out the number of molecules we multiply the moles by NA

6869.4 mol . 6.02×10²³ = 4.13×10²⁷ molecules

4 0
3 years ago
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