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salantis [7]
3 years ago
8

How many oxygen atoms are present in 0.500 mol carbon dioxide

Chemistry
1 answer:
Paha777 [63]3 years ago
6 0
Carbon dioxide = CO2

.5 mol CO2 = 1 mol O

1 mol O = <span><u>6.022140857 × 10^23  Oxygen atoms</u></span>
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R is the midpoint of FG.<br><br> FR = 26 find RG
vichka [17]

Answer:

52

Explanation:

This is because if R is the midpoint of FRG, FR is half of FRG, so basically all you do it multiply by 2 to get the FRG

7 0
3 years ago
Hat is the gram formula mass of caso4 •2h2o
Trava [24]
Your answer would be 172.1703 g/ml
7 0
3 years ago
Viscous a chemical or physical property
Ahat [919]

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chemical

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4 0
3 years ago
3.8 liters of sulfur vapor, S8(g), at 921.4°C and 5.87 atm is burned in excess pure oxygen gas to give sulfur dioxide gas measur
lilavasa [31]

Answer:

116.5 g of SO₂ are formed

Explanation:

The reaction is:

S₈(g) +  8O₂(g)  → 8SO₂ (g)

Let's identify the moles of sulfur vapor, by the Ideal Gases Law

We convert the 921.4°C to Absolute T° → 921.4°C + 273 = 1194.4 K

5.87 atm . 3.8L = n . 0.082 L.atm/mol.K . 1194.4K

(5.87 atm . 3.8L) / (0.082 L.atm/mol.K . 1194.4K) = n → 0.228 moles of S₈

Ratio is 1:8, 1 mol of sulfur vapor can produce 8 moles of dioxide

Then, 0.228 moles of S₈ must produce (0.228 . 8) /1 =  1.82 moles

We convert the moles to g → 1.82 moles . 64.06 g /1mol = 116.5 g

3 0
3 years ago
Read 2 more answers
The half-life of cesium-137 is 30 years. suppose we have a 200-mg sample.
Assoli18 [71]
a) To find  the mass after t years:

we will use this formula:

A = Ao / 2^n 

when A =the amount remaining

and Ao = the initial amount

and n = t / t(1/2)

by substitution:

∴ A = 200 mg/ 2^(t/30y)


b) Mass after 90 y :

by  using the previous formula and substitute t by 90 y

A = 200mg/ 2^(90y/30y)

∴ A = 25 mg

C) Time for 1 mg remaining:

when A= Ao/ 2^(t/t(1/2)

so, by substitution:

1 mg = 200 mg / 2^(t/30y)

∴2^(t/30y) = 200 mg  by solving for t

∴ t = 229 y 


7 0
4 years ago
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