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oksano4ka [1.4K]
3 years ago
5

17.

Chemistry
1 answer:
nataly862011 [7]3 years ago
3 0

Answer:

10 g of powered sugar

Explanation:

The powered sugar will have more surface are for the solvent to dissolve faster

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Assuming 100% dissociation, calculate the freezing point and boiling point of 3.11 m K3PO4(aq). Constants may be found here.
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2 years ago
If an object what is 30 kg of mass is moving at 10 m/s to the right how much kinetic energy does it have
ki77a [65]

Explanation:

m=30

v=10

K.E=1/2mv^2

=1/2×30×10^2

=1/2×30×100

=1500 joule

6 0
3 years ago
How many molecules are in 3 moles of NaCl?
PtichkaEL [24]

Answer:

Given : No. Of moles = 1.5

To calculate : no. Of molecules =N

We know that moles = N / 6.022 x 10²³

Therefore, 1.5 x 6.022 x 10²³ = N

Hence N = 9.0330x 10²³ molecules

3 0
3 years ago
Argon (Ar) and helium (He) are initially in separate compartments of a container at 25°C. The
love history [14]

Answer:

(a) V_B=11.68L

(b) x_{He}=0.533

Explanation:

Hello,

In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:

n_A=\frac{0.082\frac{atm*L}{mol*K}*298K}{1.974 atm*6.00L}=2.063mol

Thus, since the final pressure is 3.60 bar, we can write:

P=x_{Ar}P_A+x_{He}P_B\\\\P=\frac{n_{Ar}}{n_{Ar}+n_{He}} P_A+\frac{n_{He}}{n_{Ar}+n_{He}} P_B\\\\3.60bar=\frac{2.063mol}{2.063mol+n_{He}} *2.00bar+\frac{n_{He}}{2.063mol+n_{He}} *5.00bar

The moles of helium could be computed via solver as:

n_{He}=2.358mol

Or algebraically:

3.60bar=\frac{1}{2.063mol+n_{He}} *(4.0126+5.00*n_{He})\\\\7.314+3.60n_{He}=4.013+5.00*n_{He}\\\\7.314-4.013=5.00*n_{He}-3.60n_{He}\\\\n_{He}=\frac{3.3}{1.4}=2.358mol

In such a way, the volume of the compartment B is:

V_B=\frac{n_{He}RT}{P_B}=\frac{2.358mol*0.082\frac{atm*L}{mol*K}*298.15K}{4.935atm}\\  \\V_B=11.68L

Finally, he mole fraction of He is:

x_{He}=\frac{2.358}{2.358+2.063}\\ \\x_{He}=0.533

Regards.

8 0
3 years ago
A pure white crystalline compound was found to melt at 112.5-113.0oC when taken on a melting point apparatus, and on further hea
Sergeeva-Olga [200]

According to the question, the determined melting point of the compound is 112.5-113.0oC. When the solidified compound was retried, the melting point was found to be 133.6-154.5oC. This greater range higher than  112°C is caused by reusing samples leads to errors.

A pure sample is known by its sharp melting point. A pure sample does not melt over a large range. We can see this in the predetermined melting points of the pure sample(112.5-113.0oC).

However, reusing a sample introduces errors because the pure sample may become contaminated leading to a larger and higher range of melting point (133.6-154.5oC) which is far above  112°C.

Learn more: brainly.com/question/5325004

8 0
2 years ago
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