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Fed [463]
2 years ago
7

Hello there :3 Please anwser this Combined gas law problem for me . By the way just anwser #4 for me that’s all . I uploaded the

formulas on my other question if you need to use it . Please help me I don’t understand this please be an expert I will mark brianliest don’t scam me and SHOW ALL THE WORK! I just don’t understand *.* . Will report any links !

Chemistry
1 answer:
stellarik [79]2 years ago
6 0

Answer:

136L

Explanation:

p1v1=p2v2

114 x44.0=37.0x V2

V2=136L

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A 12.82 g sample of a compound contains 4.09 g potassium (K), 3.71 g chlorine (Cl), and oxygen (O). Calculate the empirical form
Rina8888 [55]
The  empirical  formula of the  compound is  calculated  as  follows

first   calculate  the  mass  of  oxygen=  12-(4.09  +3.71)=  5.02g

then  calculate  the  moles  of  each  element,  moles  =  mass/  molar  mass

moles of   K  =  4.09g/39 g/mol(molar  mass  of K)  =  0.105  moles
moles  of Cl = 3.71g/35.5 g/mol(molar  mass  of Cl) =  0.105  moles
moles of  O  =  5.02g/ 16g/mol(molar  mass of  O) = 0.314  moles

then  calculate e  mole ratio by  dividing  each  mole  by  the  smallest  number  of  moles  (  0.105 moles)

K=0.105/0.105= 1
Cl=0.105 /0.105=1
O=  0.314/0.105=3

therefore  the  empirical   formula  = KClO3
7 0
3 years ago
A compound is 52.0% zinc 9.6% carbon and 38.4% oxygen. Calculate the empirical formula of the compound.
skad [1K]

Answer:

So a compound is 52% Zinc(Zn), 9.6% Carbon(C), and 38.4% Oxygen (O). Let’s first start off by assuming that we have 100 g of this compound. This means that we have 52 g of Zinc, 9.6 g of Carbon, and 38.4 g of Oxygen.Zinc = 65.38 g/molCarbon = 12 g/molOxygen = 16 g/molThis means we have:52 g of Zn(1 mol Zn/65.38 g of Zn) ≈0.8 mol of Zn.9.6 g of C(1 mol C/12 g of C) = 0.8 mol of C38.4 g of O(1 mol of O/16 g of O) = 2.4 mol of O.

Explanation:

What we want to do next is divide each element by the common factor of all of them, which is 0.8. In most cases, you divide each element by the element with the least amount of moles. After we divide each by 0.8, you’ll notice you have 1 Zn, 1 C, and 3 O. This gives you the empirical formula of ZnCO3, or Zinc Carbonate.

6 0
2 years ago
There are 2 gasses, A, B. They weigh 2.46g and 0.5g respectively, and the Volume of A is 3 times the volume of B. A has a molecu
dangina [55]

Answer:

B

Explanation:

molecular mass of B is 28

8 0
2 years ago
Read 2 more answers
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 an Atom? please help
JulsSmile [24]

Answer:

An atom is the basic building block of matter. Anything that has a mass-- in other words, anything that occupies space--is composed of atoms.

5 0
3 years ago
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