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patriot [66]
3 years ago
7

Aluminum has a density of 2.70 g/mL. Calculate the mass (in grams) of a piece of aluminum having a volume of 405 mL .

Chemistry
1 answer:
bezimeni [28]3 years ago
7 0
In this question, you are given the density (2.7g/ml) and the volume of the aluminum. You are asked how much the volume is. To answer this question, you need to convert the volume into mass. Since the unit used is already same, you don't need to convert it into another unit. The calculation would be:

mass= volume / density
mass= 405ml / (2.70 g/ml)
mass= 150grams
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1. A student collected the following data for a fixed volume of gas: Temperature (⁰C) Pressure (mm of Hg) 10 726 20 750 40 800 7
zimovet [89]

Answer is: the missing pressure is 1088.66 mmHg.

Gay-Lussac's Law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.

p₁/T₁ = p₂/T₂.

p₁ = 960 mmHg; pressure of the gas.

T₁ = 100°C + 273.15.

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6 0
3 years ago
Which indicator will be the same color in hydrochloric acid as it is in pure water?
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3 0
3 years ago
If an ideal gas has a pressure of 2.97 atm, a temperature of 449 K, and has a volume of 58.35 L, how many miles of gas are in th
soldi70 [24.7K]
N = ?

T = 449 K

V = 58.35 L

P =2.97

R = 0.082

Use the clapeyron equation:

P x V = n x R x T

2.97 x 58.35 = n x 0.082 x 449

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hope this helps!



7 0
3 years ago
If you start with 6 moles of N2 and 6 moles of H2 (meaning you won't have enough of 1 of the ingredients), how many moles of NH3
Ivahew [28]

Answer:

4molNH_3

Explanation:

Hello there!

In this case, according to the given information it will be firstly necessary to set up the chemical equation taking place:

N_2+3H_3\rightarrow 2NH_3

We infer we need to calculate the moles of NH3 by using both of the moles of N2 and H2 at the beginning, in order to identify the limiting reactant:

n_{NH_3}=6molN_2*\frac{2molNH_3}{1molN_2}=12molNH_3\\\\ n_{NH_3}=6molH_2*\frac{2molNH_3}{3molH_2}=4molNH_3\\

Thus, since hydrogen yields the fewest moles of ammonia, we conclude that we are just able to yield 4 moles of NH3.

Regards!

8 0
3 years ago
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