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3 years ago

How many moles of atoms are there in each of the following?

Chemistry

1 answer:

valina [46]3 years ago

7 0

Answer:

a :1.0025

b:0.5

c:0.1

Explanation:

a :multiplicas 40.1g de ca por 1mol/masa atómica

b:multiplicas 11.5 g de Na por 1mol/masa atómica

c: multiplicas 5.87 g de Ni por 1mol/masa atómica

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A. 2.0x1024 atoms of S to moles (3.3)

Fynjy0 [20]

3.3 moles of sulfur

Explanation:

To find the number of moles knowing the number of atoms we use Avogadro's number to formulate the following reasoning:

if in 1 mole of sulfur (S) there are 6.022 × 10²³ atoms of sulfur (S)

then in X moles of sulfur (S) there are 2 × 10²⁴ atoms of sulfur (S)

X = (1 × 2 × 10²⁴) / (6.022 × 10²³)

X = 3.3 moles of sulfur

Learn more about:

Avogadro's number

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#learnwithBrainly

5 0

3 years ago

How many molecules are in 0.55 moles of Cu(NO3)2?

xeze [42]

<h3>Answer:</h3>

3.3 × 10²³ molecules Cu(NO₃)₂

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

<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>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

0.55 mol Cu(NO₃)₂

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

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 0.55 \ mol \ Cu(NO_3)_2(\frac{6.022 \cdot 10^{23} \ molecules \ Cu(NO_3)_2}{1 \ mol \ Cu(NO_3)_2})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 3.3121 \cdot 10^{23} \ molecules \ Cu(NO_3)_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

3.3121 × 10²³ molecules Cu(NO₃)₂ ≈ 3.3 × 10²³ molecules Cu(NO₃)₂

4 0

3 years ago

Most gases ______________ .

kompoz [17]

Answer is D, they are compressible

5 0

3 years ago

Look at the diagram below.

antiseptic1488 [7]

If it gains an electron it will be negatively charged and if it loses an electron it will be positively charged

7 0

3 years ago

Read 2 more answers

A mixture containing 0.477 mol he(g), 0.265 mol ne(g), and 0.115 mol ar(g) is confined in a 7.00-l vessel at 25 ∘c. part a calcu

enot [183]

Q1)
we can use the ideal gas law equation to find the total pressure of the system ;
PV = nRT
where P - pressure
V - volume - 7 x 10⁻³ m³
n - number of moles
total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in K - 273 + 25 °C = 298 K
substituting the values in the equation
P x 7 x 10⁻³ m³ = 0.857 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
P = 303.33 kPa
1 atm = 101.325 kPa
Therefore total pressure - 303.33 kPa / 101.325 kPa/atm = 2.99 atm

Q2)
partial pressure is the pressure exerted by the individual gases in the mixture.
partial pressure for each gas can be calculated by multiplying the total pressure by mole fraction of the individual gas.

total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
mole fraction of He - $\frac{0.477}{0.857} = 0.557$
mole fraction of Ne - $\frac{0.265}{0.857} = 0.309$
mole fraction of Ar - $\frac{0.115}{0.857} = 0.134$
partial pressure - total pressure x mole fraction
partial pressure of He - 2.99 atm x 0.557 = 1.67 atm
partial pressure of Ne - 2.99 atm x 0.309 = 0.924 atm
partial pressure of Ar - 2.99 atm x 0.134 = 0.401 atm

6 0

3 years ago