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Lubov Fominskaja [6]
3 years ago
9

What do atoms consist of?

Chemistry
2 answers:
kolbaska11 [484]3 years ago
7 0
Charge and uncharged particles
Mademuasel [1]3 years ago
5 0

Answer:

changed and unchanged particles

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Is volume conserved​
svlad2 [7]

Answer:

no, volume isn't conserved

6 0
3 years ago
If I contain 3 moles of gas in a container with a volume of 60 liters and at a
viktelen [127]

Answer:    n = 3.0 moles

V = 60.0 L

T = 400 K

From PV = nRT, you can find P

P = nRT/V = (3.0 mol)(0.0821 L-atm/K-mol)(400 K)/60.0L

P = 1.642 atm = 1.6 atm (to 2 significant figures)

4 0
3 years ago
Read 2 more answers
How to prepared sodium chloride solution in the laboratory.​
Afina-wow [57]

hope it will help you

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.

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hope it will helps

5 0
2 years ago
How much sucrose (g) do you need to weight in order to prepare 19.16 g of a 13.1 % (weight percent) solution?
Nonamiya [84]
<span>2.51 grams
   You want to prepare 19.16 g of some solution which will have 13.1% of it's mass being sucrose. So we just need to perform some simple multiplication: 19.16g * 0.131 = 2.50996g
   Rounding to 3 significant figures gives 2.51 g.</span>
3 0
3 years ago
What volume of oxygen (in L) is produced
sveticcg [70]

Answer:

12.36 L.

Explanation:

We'll begin by calculating the number of mole in 147.1 g of lead(II) nitrate, Pb(NO₃)₂. This can be obtained as follow:

Molar mass of Pb(NO₃)₂ = 207.2 + 2[14.01 + (16×3)]

= 207.2 + 2[14.01 + 48]

= 207.2 + 2[62.01]

= 207.2 + 124.02

= 331.22 g/mol

Mass of Pb(NO₃)₂ = 147.1 g

Mole of Pb(NO₃)₂ =?

Mole = mass / Molar mass

Mole of Pb(NO₃)₂ = 147.1 / 331.22

Mole of Pb(NO₃)₂ = 1.104 moles.

Next, we shall determine the number of mole of oxygen gas, O₂, produce from the reaction. This can be obtained as follow:

2Pb(NO₃)₂ —> 2PbO + 4NO₂ + O₂

From the balanced equation above,

2 moles of Pb(NO₃)₂ decomposed to produce 1 mole of O₂.

Therefore, 1.104 moles of Pb(NO₃)₂ will decompose to produce = (1.104 × 1)/2 = 0.552 mole of O₂.

Finally, we shall determine the volume occupied by 0.552 mole of oxygen gas, O₂. This can be obtained as follow:

1 mole of O₂ occupied 22.4 L at STP.

Therefore, 0.552 mole of O₂ will occupy = 0.552 × 22.4 = 12.36 L at STP.

Thus, the volume of oxygen gas, O₂ produced is 12.36 L.

6 0
3 years ago
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