What is the mass of 0.46 mol of MgCl2

How many molecules of ammonia are contained in 10.4 moles of ammonia, NH3?

seraphim [82]

Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. This number can be used to convert the number of atoms or molecules into number of moles. We do as follows:

10 mol NH3 ( 6.022 x 10^23 molecules / 1 mol ) = 6.022x10^24 molecules NH3

4 0

3 years ago

For a first-order reaction, A → B, the rate coefficient was found to be 3.4 × 10-4 s-1 at 23 °C. After 5.0 h, the concentration

Illusion [34]

Answer:

the original concentration of A = 0.0817092 M

Explanation:

A reaction is considered to be of first order it it strictly obeys the graphical equation method.

$k_1 = \dfrac{2.303}{t}log \dfrac{a}{a-x}$

where;

k = the specific rate coefficient = 3.4 × 10⁻⁴ s⁻¹

t = time = 5.0 h = 5.0 × 3600 = 18000 seconds

a = initial concentration = ???

a - x = remaining concentration of initial concentration at time t = 0.00018 mol L⁻¹

$3.4 \times 10^{-4}= \dfrac{2.303}{18000}log \dfrac{a}{0.00018}$

$3.4 \times 10^{-4}= 1.27944 \times 10^{-4} \times log \dfrac{a}{0.00018}$

$\dfrac{3.4 \times 10^{-4}}{1.27944 \times 10^{-4}}= log \dfrac{a}{0.00018}$

$2.657= log \dfrac{a}{0.00018}$

$10^{2.657}= \dfrac{a}{0.00018}$

$453.94 = \dfrac{a}{0.00018}$

$a =453.94 \times 0.00018$

a = 0.0817092 M

Thus , the original concentration of A = 0.0817092 M

8 0

3 years ago

What are the causes and consequences of climate change

Sati [7]

Answer: Humans are increasingly influencing the climate and the earth's temperature by burning fossil fuels, cutting down forests and farming livestock.

Explanation: This adds enormous amounts of greenhouse gases to those naturally occurring in the atmosphere, increasing the greenhouse effect and global warming.

7 0

2 years ago

CH4 + O2 → CO2 + H2O

Scilla [17]

Answer:

9.8 × 10²⁴ molecules H₂O

General Formulas and Concepts:

Atomic Structure

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

Organic

Naming carbons

Stoichiometry

Analyzing reaction rxn
Using Dimensional Analysis

Explanation:

Step 1: Define

[RxN - Unbalanced] CH₄ + O₂ → CO₂ + H₂O

[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O

[Given] 130 g CH₄

Step 2: Identify Conversions

Avogadro's Number

[RxN] 1 mol CH₄ → 2 mol H₂O

[PT] Molar Mass of C: 12.01 g/mol

[PT] Molar Mass of H: 1.01 g/mol

Molar Mass of CH₄: 12.01 + 4(1.01) = 16.05 g/mol

Step 3: Stoichiometry

[DA] Set up conversion: $\displaystyle 130 \ g \ CH_4(\frac{1 \ mol \ CH_4}{16.05 \ g \ CH_4})(\frac{2 \ mol \ H_2O}{1 \ mol \ CH_4})(\frac{6.022 \cdot 10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 9.75526 \cdot 10^{24} \ molecules \ H_2O$

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

9.75526 × 10²⁴ molecules H₂O ≈ 9.8 × 10²⁴ molecules H₂O

8 0

2 years ago

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

Answer: The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0

3 years ago