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Valentin [98]
1 year ago
6

What is the molar mass of carbon monoxide (in units of g/mol)? Use two decimal places in atomic masses. Only give the numeric va

lue of your answer.Respond with the correct number of significant figures in scientific notation (Use E notation and only 1 digit before decimal e.g. 2.5E5 for 2.5 x 10⁵)
Chemistry
1 answer:
ikadub [295]1 year ago
3 0

ANSWER

The molar mass of carbon monoxide is 28g/mol or 2.8 x 10^1 g/mol

EXPLANATION

Given that

The chemical formula of carbon monoxide is CO

Note that, the unit mass of carbon is 12.01u and the unit mass of oxygen is 15.99u

Based on the atomic masses of the elements, we can now find the molar mass of CO

\begin{gathered} \text{ CO = \lparen1 }\times\text{ 12.01\rparen + \lparen1 }\times\text{ 15.99\rparen} \\ \text{ CO = 12.01 + 15.99} \\ \text{ CO = 28g/mol} \\ \text{ CO = 2.8}\times\text{ 10}^1\text{ g/mol} \end{gathered}

Therefore, the molar mass of carbon monoxide is 28g/mol or 2.8 x 10^1 g/mol

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Answer:

Group 12

Explanation:

Group 12 transition metals are diamagnetic. They behave properties that distinguish them. They naturally have twelve electrons hence their outermost shell is fully filled.

Transition metals have high densities which increases down the group. However, the increase in density of transition elements of group 12 varies with temperature at a rate that is quite different from other transition elements. Hence the differences in the value of melting points and density changes by only a very small amount as you come down group 12 compared to other groups of transition elements.

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grandymaker [24]

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B

Explanation:

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3 years ago
a scientist uses 68 grams of CaCo3 to prepare 1.5 liters of solution. what is the molarity of this solution?
victus00 [196]

Answer: 0.4533mol/L

Explanation:

Molar Mass of CaCO3 = 40+12+(16x3) = 40+12+48 = 100g/mol

68g of CaCO3 dissolves in 1.5L of solution.

Xg of CaCO3 will dissolve in 1L i.e

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7 0
4 years ago
What pressure in atmospheres(atm) is equal to 45.6 kPa?
postnew [5]
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kolezko [41]

Answer:

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