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3 years ago

Help me with homework.

Chemistry

1 answer:

andrey2020 [161]3 years ago

7 0

The chemical equation is said to be balanced if the number of atoms in the reactants and products is the same

<h3>Further explanation</h3>

Equation balanced ⇒ total number of atoms in reactants(on the left)= total number of atoms in products(on the right)

H₂+O₂---> H₂O

Reactants : H₂, O₂

Products : H₂O

not balanced

H₂O₂ ---> H₂O+O₂

Reactants : H₂O₂

Products : H₂O, O₂

not balanced

Na+O₂ ---> Na₂O

Reactants : Na, O₂

Products : Na₂O

not balanced

N₂+H₂ ---> NH₃

Reactants : N₂, H₂

Products : NH₃

not balanced

P₄+O₂---> P₄O₁₀

Reactants : P₄, O₂

Products : P₄O₁₀

not balanced

Fe+H₂O ----> Fe₃O₄ + H₂

Reactants : Fe, H₂O

Products : Fe₃O₄

not balanced

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An element contains only one kind of A. particle. B. property. C. protein. D. raw material.

ivanzaharov [21]

Answer:

Explanation:

an element contains only one kind of particle

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3 years ago

The bromination of acetone is acid-catalyzed.CH3COCH3 + Br2 CH3COCH2Br + H+ + Br -The rate of disappearance of bromine was measu

Ann [662]

Answer:

a) The rate law is:

rate = k[Acetone][Br₂]⁰[H⁺] = k[Acetone][H⁺]

b) The value of k is:

k = 3.86 × 10⁻³ M⁻¹ · s⁻¹

Explanation:

Acetone (M) Br2 (M) H+ (M) Rate (M/s)

0.30 0.050 0.050 5.7 x 10-5

0.30 0.10 0.050 5.7 x 10-5

0.30 0.050 0.10 1.2 x 10-4

0.40 0.050 0.20 3.1 x 10-4

0.40 0.050 0.050 7.6 x 10-5

A generic rate law for this reaction could be written as follows:

rate = k[Acetone]ᵃ[Br₂]ᵇ[H⁺]ⁿ

The rate for the reaction in trial 2 is:

rate 2 = 5.7 ×10⁻⁵M/s = k(0.3)ᵃ(0.1)ᵇ(0.050)ⁿ

For the reaction in trial 1:

rate 1 = 5.7 ×10⁻⁵M/s = k(0.3)ᵃ(0.050)ᵇ(0.050)ⁿ

If we divide both expressions, we can obtain "b": rate2 / rate1:

rate2/rate1 = k(0.3)ᵃ(0.1)ᵇ(0.050)ⁿ / k(0.3)ᵃ(0.050)ᵇ(0.050)ⁿ

1 = 2ᵇ

b = 0

If we now take the expressions from trial 3 and 1 and divide them, we can obtain "n":

rate 3/rate 1 = k(0.3)ᵃ(0.050)⁰(0.01)ⁿ/ k(0.3)ᵃ(0.050)⁰(0.050)ⁿ

2.1 = 2ⁿ Applying ln to both side of the equation:

ln 2.1 = n ln2

ln2.1/ln2 = n

1 ≅ n

Taking now the reaction in trial 5 and 1 and dividing them:

rate 5/rate 1 = k(0.4)ᵃ(0.050)⁰(0.050) / k(0.3)ᵃ(0.050)⁰(0.050)

4/3 = 4/3ᵃ

a = 1

a)Then the rate law can be written as follows:

rate = k[Acetone][Br₂]⁰[H⁺]

It might be suprising that the rate of bromination of acetone does not depend on the concentration of Br₂. However, looking at the reaction mechanism, you can find out why.

b) Now, we can find the constant k for every experiment and calculate its average value:

rate / [Acetone][Br₂]⁰[H⁺] = k

For reaction 1:

k1 = 5.7 ×10⁻⁵M/s / (0.3 M)(0.050 M) = 3.8 ×10⁻³ M⁻¹ · s⁻¹

Reaction 2: k2 = 5.7 ×10⁻⁵M/s / (0.30 M)(0.050 M) = 3.8 ×10⁻³ M⁻¹ · s⁻¹

Reaction 3: k3 = 1.2 ×10⁻⁴M/s / (0.30 M)(0.10 M) = 4.0 ×10⁻³ M⁻¹ · s⁻¹

Reaction 4: k4 = 3.1 ×10⁻⁴M/s / (0.40 M)(0.20 M) = 3.9 ×10⁻³ M⁻¹ · s⁻¹

Reaction 5: k5 = 7.6 ×10⁻⁵M/s / (0.4 M)(0.05 M) = 3.8 ×10⁻³ M⁻¹ · s⁻¹

Averge value of k:

k = (k1 + k2 + k3 + k4 + k5)/5 = 3.86 × 10⁻³ M⁻¹ · s⁻¹

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3 years ago

Make propanoic anhydride from Formaldehyde

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Answer:

do you want me to make a vid and send it?

Explanation:

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2 years ago

B. if 6.73 g of na2co3 is dissolved in enough water to make 250. ml of solution, what is the molar concentration of sodium carbo

BARSIC [14]

Answer:- $[Na_2CO_3]=0.254M$ , $[Na^+]=0.508 M$ , $[CO_3^2^-]=0.254M$

Solution:- We are asked to calculate the molarity of sodium carbonate solution as well as the sodium and carbonate ions.

Molarity is moles of solute per liter of solution. We have been given with 6.73 grams of sodium carbonate and the volume of solution is 250.mL. Grams are converted to moles and mL are converted to L and finally the moles are divided by liters to get the molarity of sodium carbonate.

Molar mass of sodium carbonate is 105.99 gram per mol. The calculations for the molarity of sodium carbonate are shown below:

$\frac{6.73gNa_2CO_3}{250.mL}(\frac{1mol}{105.99g})(\frac{1000mL}{1L})$

= $0.254MNa_2CO_3$

So, molarity of sodium carbonate solution is 0.254 M.

sodium carbonate dissociate to give the ions as:

$Na_2CO_3(aq)\rightarrow 2Na^+(aq)+CO_3^2^-$

There is 1:2 mol ratio between sodium carbonate and sodium ion. So, the molarity of sodium ion will be two times of sodium carbonate molarity.

$[Na^+]=2(0.254M)$ = 0.508 M

There is 1:1 mol ratio between sodium carbonate and carbonate ion. So, the molarity of carbonate ion will be equal to the molarity of sodium carbonate.

$[CO_3^2^-]=0.254M$

5 0

3 years ago

Investigations were carried out in a science lab to explore the topic of chemical and physical changes. Investigation A Step 1.

ozzi

Answer:

Investigation B, Step 2

Explanation:

The colour change is a good indication of a chemical change. The sugar molecules were most likely changing into something else.

Investigation A, Step 1. Wrong. The dissolving of salt is a physical process.

Investigation A, Step 2. Wrong. You simply boiled off the water (a physical process) and recovered the salt.

Investigation B, Step 1. Wrong. The dissolving of sugar is a physical process.

4 0

3 years ago

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