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Pachacha [2.7K]
3 years ago
8

Initial temperature of metal= 100 C°

Chemistry
1 answer:
Vilka [71]3 years ago
3 0

Water is 22.4

Final for both us 27.1

Explanation:

I just finished this part of the lab

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If the Temperature exerted on a sample of gas is increased from 273K to 410K, and the initial pressure is 1.3317 atm what is the
dsp73

Answer: 2.00 atm

Explanation:

P1/T1 = P2/T2

1.3317/273 = P2/410

P2 = 2.00 atm

8 0
3 years ago
What is the mass of 2.3 moles of aluminum (Al)?
tangare [24]
2.3 mols of Al*26.98 molar mass of Al=62.054g
3 0
3 years ago
How is density related to the color of igneous rocks?​
r-ruslan [8.4K]

Answer:

The density of igneous rocks is related to its color. Darker colored rocks have a higher density because of its greater mineral and iron content. Its characteristics is opposite compared to lighter colored rocks that have less density because of lower mineral and iron content

4 0
3 years ago
The total concentration of ions in a 0.75 M solution of HCl is
EastWind [94]
<h3>Answer:</h3>

The total concentration of ions in a 0.75 M solution of HCl is 1.5 M

That is; 0.75 M H⁺ and 0.75 M Cl⁻

<h3>Explanation:</h3>
  • Concentration or molarity is the number of moles of a compound or an ion contained in one liter of solution. It is measured in moles per liter (M).
  • The concentration of ions making a compound is determined by the ratio of moles of the compound and the constituents ions.
  • For instance, HCl dissociates to give H⁺ and Cl⁻

       HCl(aq) → H⁺(aq) + Cl⁻(aq)

  • Therefore, since the mole ratio between HCl and the constituent ions H⁺ and Cl⁻ is 1:1, then 0.75 M of HCl dissociates to give 0.75 M H⁺ and 0.75 m Cl⁻
  • Hence the total concentration of ions in a 0.75 M solution of HCl is 1.5 M (0.75 M H⁺ and 0.75 M Cl⁻)
4 0
3 years ago
A mixture containing 20 mole % butane, 35 mole % pentane and rest
notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB

Whereas y accounts for the fractions at the outlet distillate and x for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

0.9=\frac{y_bD}{z_bF}

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

y_bD=0.9*z_bF=0.9*0.2*100mol=18mol

The total distillate flow:

y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol

And the total bottoms flow:

F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025

Then, the molar fraction of pentane and hexane:

x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422

x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

8 0
3 years ago
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