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3 years ago

14

HClO is a weak acid (Ka = 4.0 × 10–8) and so the salt NaClO acts as a weak base. What is the pH of a solution that is 0.028 M in

NaClO at 25 °C?

1 answer:

zloy xaker [14]3 years ago

7 0

This photo should help

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How many moles of sulfur<br> dioxide are in 2.26 x 10^33 sulfur dioxide molecules?

LUCKY_DIMON [66]

Answer:

moles = no. of molecules / Avogadro's number

= 2.26 x 10^33 / 6.022 x 10^23

= 3752906011

Round to significant figures which is 3 = 3.75 x 10^9 mol

Explanation: The formula for finding how many moles of a substance when given the amount of molecules is: moles = number of molecules / Avogadro's number

6 0

3 years ago

Sulfur dioxide, SO2(g), can react with oxygen to produce sulfur trioxide, SO3(g), by the following reaction

WITCHER [35]

Answer:

The heat produced is -15,1kJ

Explanation:

For the reaction:

2SO₂+O₂ → 2SO₃

The enthalpy of reaction is:

ΔHr = 2ΔHf SO₃ - 2ΔHf SO₂

As ΔHf SO₃ = -395,7kJ and ΔHf SO₂ = -296,8kJ

<em>ΔHr = -197,8kJ</em>

Using n=PV/RT, the moles of reaction are:

$n = \frac{1,00atm*3,75L}{0,082atmL/molK*298,15K}$ = <em>0,153 moles of reaction</em>

As 2 moles of reaction produce -197,8kJ of heat, 0,153moles produce:

0,153mol× $\frac{-197,8kJ}{2mol}$ = <em>-15,1kJ</em>

<em></em>

I hope it helps!

8 0

3 years ago

Oxygen, with an atomic number of 8, is a neutral atom and would have ___________ electrons in the first electron shell and _____

motikmotik

2 in the first shell and 6 in the second

8 0

3 years ago

How many moles are in 8.63 x 103 atoms of Li?

Ira Lisetskai [31]

<h3>Answer:</h3>

1.43 × 10⁻²⁰ mol Li

<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>Chemistry</u>

<u>Atomic Structure</u>

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

<h3>Explanation:</h3>

<u>Step 1: Define</u>

8.63 × 10³ atoms Li

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

Avogadro's Number

<u>Step 3: Convert</u>

Set up: $\displaystyle 8.63 \cdot 10^3 \ atoms \ Li(\frac{1 \ mol \ Li}{6.022 \cdot 10^{23} \ atoms \ Li})$
Multiply/Divide: $\displaystyle 1.43355 \cdot 10^{-20} \ moles \ Li$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.43355 × 10⁻²⁰ mol Li ≈ 1.43 × 10⁻²⁰ mol Li

4 0

2 years ago

Consider the reaction. 2 Pb ( s ) + O 2 ( g ) ⟶ 2 PbO ( s ) An excess of oxygen reacts with 451.4 g of lead, forming 367.5 g of

Art [367]

Answer : The percent yield of the reaction is, 75.6 %

Solution : Given,

Mass of Pb = 451.4 g

Molar mass of Pb = 207 g/mole

Molar mass of PbO = 223 g/mole

First we have to calculate the moles of Pb.

$\text{ Moles of }Pb=\frac{\text{ Mass of }Pb}{\text{ Molar mass of }Pb}=\frac{451.4g}{207g/mole}=2.18moles$

Now we have to calculate the moles of $PbO$

The balanced chemical reaction is,

$2Pb(s)+O_2(g)\rightarrow 2PbO(s)$

From the reaction, we conclude that

As, 2 mole of $Pb$ react to give 2 mole of $PbO$

So, 2.18 mole of $Pb$ react to give 2.18 mole of $PbO$

Now we have to calculate the mass of $PbO$

$\text{ Mass of }PbO=\text{ Moles of }PbO\times \text{ Molar mass of }PbO$

$\text{ Mass of }PbO=(2.18moles)\times (223g/mole)=486.1g$

Theoretical yield of $PbO$ = 486.1 g

Experimental yield of $PbO$ = 367.5 g

Now we have to calculate the percent yield of the reaction.

$\% \text{ yield of the reaction}=\frac{\text{ Experimental yield of }PbO}{\text{ Theoretical yield of }PbO}\times 100$

$\% \text{ yield of the reaction}=\frac{367.5g}{486.1g}\times 100=75.6\%$

Therefore, the percent yield of the reaction is, 75.6 %

3 0

3 years ago