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4 years ago

How many moles are in 5g of Na2CO3? a) 0.047b) 1c) 2d)21.2

Chemistry

1 answer:

stepladder [879]4 years ago

5 0

Answer:

0.047

Explanation:

Data : 5g , Na2CO3

Formula:

Molecular mass of Na2Co3 = no.of atomic mass x no. of atom

moles = mass÷ molecular mass

molecular mass = 2(23)+1(12)+3(16)

=106amu

moles = 5÷106

= 0.047 Ans:

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28.5 g of iron shot is added to a graduated cylinder containing 45.50 mL of water. The water level rises to the 49.10 mL mark. F

djyliett [7]

Answer:

Density if iron shot = 7.92 g/ml (2 sig. figs.)

Explanation:

By definition, density is mass per unit volume. That is,

Density (D) = mass(m)/Volume(V)

mass(m) = 28.5g (given)

Volume(V, of iron shot) = 49.10ml - 45.5ml = 3.6ml

Substitutuing into D = m/V = 28.5g/3.6ml = 7.9166667 g/ml (calc.ans)

≅ 7.92 g/ml (2 sig. figs.)

Hope this helps. :-)

7 0

3 years ago

How many atoms are in a 158.69 g sample of nickel?

Alchen [17]

<h3>Answer:</h3>

1.6283 × 10²⁴ atoms Ni

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Use Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

158.69 g Ni

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of Ni - 58.69 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 158.69 \ g \ Ni(\frac{1 \ mol \ Ni}{58.69 \ g \ Ni})(\frac{6.022 \cdot 10^{23} \ atoms \ Ni}{1 \ mol \ Ni})$
Multiply/Divide: $\displaystyle 1.62827 \cdot 10^{24} \ atoms \ Ni$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 5 sig figs.</em>

1.62827 × 10²⁴ atoms Ni ≈ 1.6283 × 10²⁴ atoms Ni

8 0

3 years ago

At the beginning of the reaction: N2 + 3H2 -> 2NH there is 1 mol of N. and 3 moles of H, and at the end of it is present a mi

Nikolay [14]

Answer:

75%

Explanation:

call realistic moles of N2 in the reaction is a

moles of H2 :3a

moles of NH3: 2a

the moles of residual N2 after reaction: 1-a

the moles of residual H2 after react: 3- 3a

the moles of mixture: 2a+1-a+3-3a=2.5

-> a=0.75

H=0.75/1=75%

7 0

3 years ago

Not really good at rounding lol please help ASAP !!!!!

stepan [7]

We need to round to the ones place (which is a 7). We look to the digit to the right of the 7 and it is a number 5 or bigger (6), so we round the 7 up one to 8.
498 g

7 0

3 years ago

If a particle had four atomic mass unit, two of which were neutrons, but had no electrons, what would the particle be?

kogti [31]

Answer:

ALPHA PARTICLE => Helium Nucleus (He⁺²)

Explanation:

3 0

3 years ago

How many moles are in 5g of Na2CO3? a) 0.047b) 1c) 2d)21.2​

How many moles are in 5g of Na2CO3? a) 0.047b) 1c) 2d)21.2