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Vitek1552 [10]
3 years ago
15

What is the maximum most that can be extracted from 76 g og Cr2O3.

Chemistry
1 answer:
tiny-mole [99]3 years ago
7 0

Answer:

D) 152 g

Explanation:

a mass of 1 mol Cr2O3 = 2 × 52 + 3 × 16 = 152 g number of moles of Cr2O3.

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Read “The Ozone Hole” and answer the question below. List at least three scientific disciplines related to chemistry mentioned o
kirill115 [55]

Answer:

Choose any of these and it should be correct:

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Explanation:

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4 0
3 years ago
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What is the molar concentration of Cu2+ in a solution that is labelled 5 ppm Cu2+? a)7.9x10^(-5)M b)7.9x10^(-6)M c)5.1x10^(-6)M
Savatey [412]

Answer:

1.24x10⁻⁴ mol/L

Something went wrong with the choices

Explanation:

5 ppm is a sort of concentration that indicates:

weigh of solute x 10⁶ / weigh of solution or volume of solution.

Relation must be in the order of 10⁻⁶, for example mg/kg / μg/g.

We can also write ppm as μg/mL, so 5 ppm will be understood as 5 μg of Ca²⁺ in 1mL of solution.

This would be the rule of three to reach molarity

In 1mL we have 5 μg of Ca²⁺

In 1000 mL we would have 5000 μg of Ca²⁺

Let's convert 5000 μg to g ( 1g = 1x10⁶μg)

5000 μg =  5x10⁻³ g

Now, that we have the mass, we convert it to moles (mass / molar mass)

5x10⁻³ g / 40.08g/m = 1.24x10⁻⁴ moles

As this moles are in 1000mL (1L) it's molarity

5 0
3 years ago
If you add a large piece of zinc to a beaker of acid and, in a separate beaker, zinc shavings to a beaker of acid (of equal conc
Sergeu [11.5K]

Answer: surface area

Explanation:

Surface area affects rate of reaction. The smaller the surface area the faster the rate of reaction. The zinc shavings have a smaller surface area of reactants compared to the large piece of zinc. The smaller surfaces area of reactants ensures that it comes into close proximity with the acid solution for reaction to take place.

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3 years ago
What is the general formula for a straight-chain alkane?
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Answer:

The general formula for a straight chain alkane is CnH(2n+...so correct option is C and D

6 0
2 years ago
How many ML of 0.44 M HCl are needed to dissolve 9.83 g of BaCO3
Tanzania [10]

Answer:

nBACO3=m/M=9,83/197=0,05(mol)  ->nHCl=0,05.2/1=0,1(mol) =>VHCl=n/CM=0,1/0,44=0,227(lít)

Explanation:

6 0
3 years ago
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