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Klio2033 [76]
3 years ago
12

What five SI base units are commonly used in chemistry?

Chemistry
2 answers:
Stells [14]3 years ago
8 0
The five most commonly used are:
•Meter
•Kilometer
•Kelvin
•Second
•Mole
Rashid [163]3 years ago
3 0
Five base SI units that are commonly used in Chemistry; are 
1.Metre (m)
2. Kilogram(kg)
3. seconds(s)
4.Kelvins (k)
5. Moles (mol)
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How do you calculate the number of moles in CO2
gogolik [260]

Answer:

The number of molecules in a mole (known as Avogadro's constant) is defined such that the mass of one mole of a substance, expressed in grams, is equal to the mean molecular mass of the substance. The molecular mass of CO2 = 12+2x16 = 44, so the mass of a mole of CO2 is approximalty 44 grams

Explanation:

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3 years ago
Help with Organic chemistry mechanism
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3H + 3Br = HBr9 Organic chemistry mechanism
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A cylinder container is filled with air. Its radius is 6.00 cm and
Ede4ka [16]

Answer:

395.84cm³

Explanation:

V= π r² h

V= 3.14 6² 3.5

V= 3.14 36 3.5

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7 0
3 years ago
Convert 18.9 moles to MgCl2 to formula units
Masja [62]

Answer:

18.9 moles of MgCl2 = 17.834 kg of MgCl2

Explanation:

The molecular weight of MgCl is 80.0 g/mol . So, to convert the given mole amount to grams, multiply this by this number, which is constant for all compounds with a specific composition (mass fraction).

Considering the original question was in the context of chemistry, I wanted to make it seem formal and more educational too. Hopefully that worked!  

EDIT: Came up with some text around what happens inside cells that would have made it better if someone just had an issue converting units, but I doubt my answer will be accepted >.<

3 0
3 years ago
The rate of effusion of an unknown gas was measured and found to be 11.9 mL/min. Under identical conditions, the rate of effusio
iren2701 [21]

Answer : The correct option is, (B) CO_2

Solution :

According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.

R\propto \sqrt{\frac{1}{M}}

or,

(\frac{R_1}{R_2})=\sqrt{\frac{M_2}{M_1}}       ..........(1)

where,

R_1 = rate of effusion of unknown gas = 11.9\text{ mL }min^{-1}

R_2 = rate of effusion of oxygen gas = 14.0\text{ mL }min^{-1}

M_1 = molar mass of unknown gas  = ?

M_2 = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the above formula 1, we get:

(\frac{11.9\text{ mL }min^{-1}}{14.0\text{ mL }min^{-1}})=\sqrt{\frac{32g/mole}{M_1}}

M_1=44.2g/mole

The unknown gas could be carbon dioxide (CO_2) that has approximately 44 g/mole of molar mass.

Thus, the unknown gas could be carbon dioxide (CO_2)

5 0
4 years ago
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