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3 years ago

PLEASE HELP ITS DUE IN 1 HOUR!!!!!!

1 answer:

8 0

Answer: Sorry I try to download it but it not showing anything, if you take a picture of the work maybe I would do it....

Explanation:

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I need help with #2. I don't know what law to use for problem.

bagirrra123 [75]

This uses something called the combined gas law. The combined gas law is as follows: (P1*V1/T1) = (P2*V2/T2)
According to question 2, you are given the following values initially:
P1 = 680 mm Hg * (1 atm/760 mm Hg) = 0.895 atm
V1 = 20.0 L
T1 = 293 K
STP or standard temperature and pressure implies that the other values we know are:
P2 = 1 atm
T2 = 273 K
Our unknown is V2
If we plug in our known values into the combined gas law:
(P1*V1/T1) = (P2*V2/T2)
(0.895 atm * 20.0 L)/293K = (1 atm * X liters)/273 K
0.0611 L*atm/K = (1 atm * X liters)/273 K
16.7 L = X liters
Therefore, the volume occupied at STP is 16.7 liters
This makes sense because the gas would occupy a smaller volume at a lower temperature, since the gas would have a lower average kinetic energy.

4 0

3 years ago

Fe2O3 + 2Al -> Al2O3 + 2Fe

pashok25 [27]

Fe2O3 + 2Al ---> Al2O3 + 2Fe
Mole ratio Fe2O3 : Al = 1:2
No. of moles of Fe2O3 = Mass/RMM = 250 / (55.8 * 2 + 16 * 3) = 1.56641604 moles
No. of moles of Al = 150/27 = 5.555555555 moles.
Mole ratio 1 : 2. 1.56641604 * 2 = 3.13283208 moles of Al, but you have 5.555555555 moles of Al. So Al is in excess. All of it won't react.

So take the Fe2O3 and Fe ratio to calculate the mass of iron metal that can be prepared.
RMM of Fe2O3 / Mass of Fe2O3 = RMM of 2Fe / Mass of Fe 159.6 / 250 = 111.6 / x x = 174.8 g of Fe

7 0

3 years ago

Read 2 more answers

For hot vacuum filtration, the filter paper should be completely dry when pouring the hot solution into the Buchner funnel to fi

Oliga [24]

Answer:

False.

Explanation:

The given statement is false because for hot vacuum filtration, the filter paper should be wet rather than dry when pouring the hot solution into the Buchner funnel. This is because The possible explanation the filter paper needs to be wetted is not only to allow it to adhere to the funnel, but also to promote the solute to filter down readily across its pores of the paper without wetting it (this is true for organic and aqueous solvents).

5 0

3 years ago

Which of the following is the best explanation for a covalent bond?

Scrat [10]

Answer:

A. electrons simultaneously attracted by more than one nucleus

Explanation:

Covalent bond is the bond which is formed with the sharing of the electrons between the two atoms which are taking part in the bond. It is generally formed between the atoms with similar electronegativity values.
It is the bond which is generally occurs within non metals as they share electrons to complete their octet.
The difference in the electronegativity values of the atoms involving in a covalent bond must not exceed the value of 1.7 .

Thus, the electrons are attracted by the two different nucleus of the atoms that are taking part in the bonding.

So, the correct answer is:- A. electrons simultaneously attracted by more than one nucleus

6 0

3 years ago

As an FDA physiologist, you need 0.625 L of phosphoric acid acid / dihydrogen phosphate (H3PO4 (aq) / H2PO4 - (aq) ) buffer with

aksik [14]

Answer:

0.4058L of 1.0M H3PO4

0.2192L of 1.5M NaOH

Explanation:

The pKa of the H3PO4 / H2PO4- buffer is 2.12

To solve this question we must use H-H equation for this system:

pH = pKa + log [H2PO4-] / [H3PO4]

2.75 = 2.12 + log [H2PO4-] / [H3PO4]

0.63 = log [H2PO4-] / [H3PO4]

4.2658 = [H2PO4-] / [H3PO4] (1)

Where [] could be taken as the moles of each reactant

As you have H3PO4 solution, the reaction with NaOH is:

H3PO4 + NaOH → H2PO4- + Na+ + H2O

As you can see, both H3PO4 and H2PO4- comes from the same 1.0M H3PO4 solution

The moles of H3PO4 are:

[H3PO4] = Moles H3PO4 - Moles NaOH

And for H2PO4-:

[H2PO4-] = Moles NaOH added

Replacing in (1):

4.2658 = [Moles NaOH] / [Moles H3PO4 - Moles NaOH]

4.2658 Moles H3PO4 - 4.2658 moles NaOH = Moles NaOH

4.2658 Moles H3PO4 = 5.2658 moles NaOH (1)

In volume:

0.625L = Moles H3PO4 / 1.0M + Moles NaOH / 1.5M

0.625 = Mol H3PO4 + 0.6667 Moles NaOH (2)

Replacing (2) in (1):

4.2658 Moles H3PO4 = 5.2658 (0.625 - Mol H3PO4 / 0.6667)

4.2658 Moles H3PO4 = 5.2658 (0.625 - Mol H3PO4) / 0.6667

4.2658 Moles H3PO4 = 5.2658*(0.9375 - 1.5 mol H3PO4)

4.2658 Moles H3PO4 = 4.9367 -7.8983 mol H3PO4

12.1641 mol H3PO4 = 4.9367

Mol H3PO4 = 0.4058moles * (1L / 1.0moles) =

0.4058L of 1.0M H3PO4

And:

0.625L - 0.4058L =

0.2192L of 1.5M NaOH

5 0

3 years ago