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motikmotik
3 years ago
13

Calculate the mass (in grams) of 8.56 moles of sulfur.

Chemistry
1 answer:
oksano4ka [1.4K]3 years ago
5 0

Answer:

275g

Explanation:

Depending on the molar mass you are given, you can use that to solve this.

(I'm going based on my science class' molar mass of sulphur being 32.07g/mol)

Starting off, the formula for finding moles is

n=m/M (moles = mass / molar mass)

We can manipulate this equation to solve for mass.

m=Mn

now fill in what we now.

m = 32.07*8.56

mass = 274.5192

Now round for significant digits (if you are needed to do)

mass = 275g

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6 0
3 years ago
Read 2 more answers
0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati
erastova [34]

Answer:

[Ag^{+}]=4.2\times 10^{-2}M

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

[K2CrO4] = 0.10 M

Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

BaCrO_4 (s)\leftrightharpoons  Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ba^{2+}][CrO_{4}^{2-}]

1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]

[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}

Now,

Ag_{2}CrO_4(s) \leftrightharpoons  2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]

1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})

[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}

[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

                       

7 0
3 years ago
How many grams is 4.2X10^24 atoms of sulfur?
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2 years ago
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A 2.5 g sample of french fries is placed in a calorimeter with 500.0 g of water at an initial temperature of 21 °C. After combus
SIZIF [17.4K]
Q=m°C<span>ΔT
=(500g) x (1 cal/g.</span>°C) x (48°C-21°C) = 13500 cal
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