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sergey [27]
3 years ago
15

What is the density of a gas with a volume of 5.6 FL oz. and a mass of 37 grams?

Chemistry
2 answers:
SIZIF [17.4K]3 years ago
7 0

Answer:

ion no

Explanation:

Assoli18 [71]3 years ago
4 0

Answer:

6.60

Explanation:

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How many grams of aluminum are required to react with 35 mL of 2.0 M hydrochloric acid, HCl?__ HCl + __ Al --> __ AlCl3 + __
kiruha [24]

Answer:

0.6258 g

Explanation:

To determine the number grams of aluminum in the above reaction;

  • determine the number of moles of HCl
  • determine the mole ratio,
  • use the mole ratio to calculate the number of moles of aluminum.
  • use RFM of Aluminum to determine the grams required.

<u>Moles </u><u>of </u><u>HCl</u>

35 mL of 2.0 M HCl

2 moles of HCl is contained in 1000 mL

x moles of HCl is contained in 35 mL

x \: mol \:  =  \:  \frac{2 \:  \times  \: 35}{1000}  \\  = 0.07 \: moles \:

We have 0.07 moles of HCl.

<u>Mole </u><u>ratio</u>

  • Balanced equation

6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)

Hence mole ratio = 6 : 2 (HCl : Al

  • but moles of HCl is 0.07, therefore the moles of Al;

=  \frac{2}{6}  \times 0.07 \\  \:  = 0.0233333 \: moles

Therefore we have 0.0233333 moles of aluminum.

<u>Grams of </u><u>Aluminum</u>

We use the formula;

grams \:  = moles \:  \times  \: rfm

The RFM (Relative formula mass) of aluminum is 26.982g/mol.

Substitute values into the formula;

= 0.0233333 \: moles  \:  \times  \: 26.982 \:  \frac{g}{mol}  \\  = 0.625799 \: grams

The number of grams of aluminum required to react with HCl is 0.6258 g.

3 0
2 years ago
What is the affect of this
Alisiya [41]

Answer:

Explanation:

Your answer should be in the attached pdf

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5 0
3 years ago
Read 2 more answers
How do chemicals in a plant relate to bath bombs?
OLga [1]

When a bath bomb comes in contact with water, the baking soda and citric acid react to make carbon dioxide bubbles. This is an acid–base reaction, where baking soda (also called sodium bicarbonate) is a weak base and citric acid is a weak acid.

3 0
2 years ago
A chemist dissolved crystals of an unknown substance into water at room temperature. He found that 33 g of the substance can be
Inga [223]
The answer is: It's solubility
8 0
3 years ago
Read 2 more answers
A geochemist in the field takes a 36.0 mL sample of water from a rock pool lined with crystals of a certain mineral compound X.
NeTakaya

Answer:

solubility of X in water at 17.0 ^{0}\textrm{C} is 0.11 g/mL.

Explanation:

Yes, the solubility of X in water at 17.0 ^{0}\textrm{C} can be calculated using the information given.

Let's assume solubility of X in water at 17.0 ^{0}\textrm{C} is y g/mL

The geochemist ultimately got 3.96 g of crystals of X after evaporating the diluted solution made by diluting the 36.0 mL of stock solution.

So, solubility of X in 1 mL of water = y g

Hence, solubility of X in 36.0 mL of water = 36y g

So, 36y = 3.96

   or, y = \frac{3.96}{36} = 0.11

Hence solubility of X in water at 17.0 ^{0}\textrm{C} is 0.11 g/mL.

3 0
3 years ago
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