All categories

3 years ago

PLEASE HELP MEEEEE (using 98 points to post)

Chemistry

2 answers:

Alla [95]3 years ago

8 0

29.2 g/mol is your answer

Gekata [30.6K]3 years ago

7 0

Δt = i Kf m

2.86 °C = (1) (1.86 °C kg mol-1) (x / 0.750 kg)

2.86 °C = (2.48 °C mol-1) (x)

x = 1.1532 mol

33.7 g / 1.1532 mol = 29.2 g/mol

You might be interested in

Ameaba is which protozoan

Vikki [24]

Answer: genus

Explanation:

Amoebas do not form a single taxonomic group; instead, they are found in every major lineage of eukaryotic organisms. Amoeboid cells occur not only among the protozoa, but also in fungi, algae, and animals.

4 0

4 years ago

Read 2 more answers

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔

Vilka [71]

Answer : The correct option is, (C) 1.1

Solution : Given,

Initial moles of $N_2O_4$ = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The given equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initially c 0

At equilibrium $(c-c\alpha)$ $2c\alpha$

The expression of $K_c$ will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

where,

$\alpha$ = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

$K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}$

$K_c=1.066\aprrox 1.1$

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0

3 years ago

When thermal energy is removed from matter, what happens to the molecules?

Colt1911 [192]

Answer:

Figure it out.

Explanation:

3 0

3 years ago

Please help..... who is up for a zoom to help me with balancing equations

Alex787 [66]

Answer:

Um I can help but not on zo.om. I can help u on here just post ur question and I will answer it

8 0

3 years ago

Read 2 more answers

How many moles are in a 12.0 g sample of NiC12

Nady [450]

Answer:

0.17 moles

Explanation:

In the elements of the periodic table, the atomic mass = molar mass. <u>Ex:</u> Atomic mass of Carbon is 12.01 amu which means molar mass of Carbon is also 12.01g/mol.

In order to find the # of moles in a 12 g sample of NiC-12, we will need to multiply the number of each atom by its molar mass and then add the masses of both Nickel and C-12 found in the periodic table:

Molar Mass of Ni (Nickel): 58.69 g/mol

Molar Mass of C (Carbon): 12.01 g/mol

Since there's just one atom of both Carbon and Nickel, we just add up the masses to find the molar mass of the whole compound of NiC-12.

58.69 g/mol of Nickel + 12.01 g/mol of Carbon = 70.7 g/mol of NiC-12

There's 12g of NiC-12, which is less than the molar mass of NiC-12, so the number of moles should be less than 1. In order to find the # of moles in NiC-12, we need to do some dimensional analysis:

12g NiC-12 (1 mol of NiC-12/70.7g NiC-12) = 0.17 mol of NiC-12
The grams cancel, leaving us with moles of NiC-12, so the answer is 0.17 moles of NiC-12 in a 12 g sample.

<em>P.S. C-12 or C12 just means that the Carbon atom has an atomic mass of 12amu and a molar mass of 12g/mol, or just regular carbon.</em>

5 0

3 years ago