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3 years ago

What is the molecular formula of a compound whose molar mass is 112 and whose empirical formula is ?

1 answer:

7 0

Well I need to know the empirical formula, molecular formula.
Then I can multiply the grams of each individual element by their molar mass, and divide the smallest element by itself, and the other elements by the smallest element. Then I get the empirical formula

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The pH of an acidic solution is 2.11. What is [H⁺]

BartSMP [9]

Answer:

denotes the molar hydrogen ion concentration

Explanation:

3 0

3 years ago

Read 2 more answers

A sample of calcium phosphate was found to have a mass of 125.3 g. How many molecules were contained in the sample?

Viktor [21]

The answer for the following problem is mentioned below.

Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules.

Explanation:

Given:

mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 125.3 grams

We know;

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = (40×3) + 3 (31 +(4×16))

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 120 + 3(95)

molar mass of calcium phosphate ( $Ca_{3}(PO_{4} )_{2}$ ) = 120 +285 = 405 grams

We also know;

No of molecules at STP conditions( $N_{A}$ ) = 6.023 × 10^23 molecules

To solve:

no of molecules present in the sample(N)

We know;

$\frac{m}{M} =\frac{N} }{}$ N÷ $N_{A}$

$\frac{405}{125.3}$ = $\frac{N}{6.023*10^23}$

N =(405×6.023 × 10^23) ÷ 125.3

N = 19.3 × 10^23 molecules

Therefore number of molecules(N) present in the calcium phosphate sample are 19.3 × 10^23 molecules

3 0

3 years ago

How do you Calculate the molarity of 250 ml of solution in which 2.7 g of MgCl2 are dissolved

IRINA_888 [86]

Answer:

Molar mass of MgCl2 is 95 g/mol

Mg = 24 g/mol and Cl = 35.5 ×2 = 71 g/mol

moles = mass given/ molar mass

= 2.7/95 = 0.028 mol

volume = 250/1000 = 0.25 dm3 (ml is the same as dm3)

molarity of MgCl2 = moles/volume

= 0.028/0.25

= 0.112 mol/dm3

3 0

2 years ago

to find the mass of 3.2 moles of calcium for carbonate which of the following is used to calculate the amount

natka813 [3]

I believe it to be g/mol of Calcium carbonate
because to finde mass...u must have grams(g) as units....
it is the only one that have g as units

as for the first answer the avogadros number gives u the number of atoms in one mole of calcium carbonate....
the second one is based on ... At s.t.p one mole of gas occupies 22.4 dm³⇒to find volume

6 0

2 years ago

What is the mass of 3.77mol of K3N?

AleksAgata [21]

<h3>Answer:</h3>

495 g K₃N

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Reading a Periodic Table

Stoichiometry

Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

3.77 mol K₃N

Step 2: Identify Conversions

Molar Mass of K - 39.10 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of K₃N - 3(39.10) + 14.01 = 131.31 g/mol

Step 3: Convert

Set up: $\displaystyle 3.77 \ mol \ K_3N(\frac{131.31 \ g \ K_3N}{1 \ mol \ K_3N})$
Multiply/Divide: $\displaystyle 495.039 \ g \ K_3N$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

495.039 g K₃N ≈ 495 g K₃N

5 0

3 years ago