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3 years ago

The activation energy of a chemical reaction is the energy that

Chemistry

1 answer:

Svetlanka [38]3 years ago

3 0

Explanation:

The activation energy of a chemical reaction is the amount of energy that must be added to go from the energy level of the reactants to the energy level of transition state.

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Write ionic equations for any three of the following:

worty [1.4K]

The ionic equation is:

H⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + NO₃⁻(aq) + H₂O(l)

HNO₃(aq) + NaOH(aq) → NaNO₃(aq) + H2O(l)

The required dissociations are

HNO₃(aq) → H⁺(aq) + NO₃⁻(aq)

NaOH(aq) → Na⁺(aq) + OH⁻(aq)

NaNO₃(aq) → Na⁺(aq) + NO₃⁻(aq)

So, the ionic equation is

H⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + NO₃⁻(aq) + H₂O(l)

The ionic equation is

H⁺(aq) + Cl⁻(aq) + K⁺(aq) + OH⁻(aq) → K⁺(aq) + Cl⁻(aq) + H₂O(l)

HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)

The required dissociations are

HCl(aq) → H⁺(aq) + Cl⁻(aq)

KOH(aq) → K⁺(aq) + OH⁻(aq)

KCl(aq) → K⁺(aq) + Cl⁻(aq)

So, the ionic equation is

H⁺(aq) + Cl⁻(aq) + K⁺(aq) + OH⁻(aq) → K⁺(aq) + Cl⁻(aq) + H₂O(l)

The ionic equation is

2H⁺(aq) + SO₄²⁻(aq) + Mg²⁺(aq) + 2OH⁻(aq) → Mg²⁺(aq) + SO₄²⁻(aq)

+ 2H₂O(l)

H₂SO₄(aq) + Mg(OH)₂(aq) → MgSO₄(aq) + 2H2O(l)

The required dissociations are

H₂SO₄(aq) → 2H⁺(aq) + SO₄²⁻(aq)

Mg(OH)₂(aq) → Mg²⁺(aq) + 2OH⁻(aq)

MgSO₄(aq) → Mg²⁺(aq) + SO₄²⁻(aq)

So, the ionic equation is

2H⁺(aq) + SO₄²⁻(aq) + Mg²⁺(aq) + 2OH⁻(aq) → Mg²⁺(aq) + SO₄²⁻(aq)

+ 2H₂O(l)

Learn more about ionic equations here:

brainly.com/question/11628165

5 0

3 years ago

How many nanometers are in 0.0006245101 km?

maw [93]

Answer:

624510100

Explanation:

Doing a conversion factor:

$0,0006245101[km]*\frac{1000[m]}{1 km} *\frac{1x10^{9} nanometer}{1 m} =624510100 [nanometer]$

5 0

3 years ago

What does it mean ppppllllllzzzzz

Dafna1 [17]

Method means a particular form or procedure for accomplishment or approaching something, especially a systematic or established one.

4 0

3 years ago

Read 2 more answers

Calculate hydrochloric acid (umol)in 200 ul of a 0.5173Msolution of acid?

strojnjashka [21]

Answer: The moles of hydrochloric acid is $1.0346\times 10^{-4}\mu mol$

Explanation:

To calculate the molarity of solution, we use the equation:

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

Or,

$\text{Molarity of the solution}=\frac{\text{Micro moles of solute}\times 10^6}{\text{Volume of solution (in }\mu L)}}$

We are given:

Molarity of solution = 0.5173 M

Volume of solution = $200\mu L$

Putting values in above equation, we get:

$0.5173M=\frac{\text{Micro moles of HCl}\times 10^6}{200\mu L}\\\\\text{Micro moles of HCl}=1.0346\times 10^{-4}\mu mol$

Hence, the moles of hydrochloric acid is $1.0346\times 10^{-4}\mu mol$

7 0

3 years ago

Suppose 0.795 g of sodium iodide is dissolved in 100. mL of a 39.0 m M aqueous solution of silver nitrate.

lubasha [3.4K]

Answer:

The final molarity of iodide anion is 0.053 M

Explanation:

Step 1: Data given

Mass of sodium iodide (NaI) = 0.795 grams

Volume of the solution = 100 mL = 0.1 L

Molarity of aqueous solution of silver nitrate (AgNO3) = 39 mM = 0.039M

The molecular mass of sodium iodide is 149.89 g/mol.

Step 2: The balanced equation

AgNO3(aq) + NaI(aq) → AgI(s) + NaNO3(aq)

Step 3: Calculate number of moles of sodium iodide

Moles NaI = mass NaI / Molar mass NaI

Moles NaI = 0.795 grams / 149.89 g/mol

Moles NaI = 0.0053 moles

For 1 mole AgNO3 consumed, we need 1 mole NaI to produce 1 mole AgI and 1 mole NaNO3

The sodium iodide will dissociate as followed:

NaI(aq) → Na+(aq) + I-(aq)

Step 4: Calculate iodide ions

For 1 mole NaI, we have 1 mole of I-

For 0.0053 moles of NaI we'll have 0.0053 moles I-

Step 5: Calculate molarity of iodide ion

Molarity = moles I- / volume

Molarity I- = 0.0053 moles / 0.1 L

Molarity I- = 0.053 M

The final molarity of iodide anion is 0.053 M

5 0

3 years ago