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3 years ago

BRAINLIESTTT ASAP!! PLEASE HELP ME ASAP!

Chemistry

1 answer:

Artyom0805 [142]3 years ago

7 0

Answer:

Here's what I get.

Explanation:

There are many ways of classifying a reaction. Here are some that can apply to yours.

A. Combination or synthesis

Two or more elements combine to form a single compound.

B. Redox or reduction/oxidation

One element is oxidized, and another is reduced.

For example, the oxidation number of N is reduced from 0 to -3, and that of H is increased from 0 to +1.

C. Gas phase reaction

All reactants and products are gases (although the reaction is extremely slow without a solid catalyst).

D. Homogeneous

All substances are in the same phase (if you aren’t using a solid catalyst).

The main types of reaction are

Combination

Decomposition
Single displacement
Double displacement
Combustion
Redox

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2. A chemical analysis of a sample provides the following elemental data:

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Answer:

C3 H6 O2

Explanation:

first divide their mass by their respective molar mass, we get:

30.4 moles of C

61.2 moles of H

20.25 moles of O

now divide everyone by the smallest one of them then we get

C= 1.5

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since our answer of C is not near to any whole number so we will multiply all of them by 2

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C3 H6 O2 is our answer

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Read 2 more answers

The rate of a certain reaction is given by the following rate law: rate Use this information to answer the question below. At a

AleksAgata [21]

Answer:

Initial rate of the reaction when concentration of hydrogen gas is doubled will be $3.2\times 10^6 M/s$ .

Explanation:

$N_2+3H_2\rightarrow 2NH_3$

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

Initial rate of the reaction = R = $4.0\times 10^5 M/s$

$R = k\times [N_2][H_2]^3$

$4.0\times 10^5 M/s=k\times [N_2][H_2]^3$

The initial rate of the reaction when concentration of hydrogen gas is doubled : R'

$[H_2]'=2[H_2]$

$R'=k\times [N_2][H_2]'^3=k\times [N_2][2H_2]^3$

$R'=8\times k\times [N_2][H_2]^3$

$R'=8\times R=8\times 4\times 10^5 M/s=3.2\times 10^6 M/s$

Initial rate of the reaction when concentration of hydrogen gas is doubled will be $3.2\times 10^6 M/s$ .

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3 years ago

calculate how much acid (acetic acid) and how much conjugate base (sodium acetate) must be used to make 500ml of a 0.8m acetate

kirza4 [7]

For the desired pH of 5.76, 0.365 mol of acetate and 0.035 mol of acid are to be added

let the concentration of acetate be x

then the concentration of acid will be (0.8 - x)

pKa of acetate buffer = 4.76

pH = pKa + log([acetate]/[acid])

⇒4.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 0

⇒x/(0.8-x) = 1

⇒x = 0.4

Therefore

[acetate] = x = 0.4

[acid] = 0.8-x =0.4 M

number of mol = concentration *(volume in mL)

number of mol of acetate = 0.4*0.5

= 0.20 mol

number of mol acid = 0.4*0.5

= 0.20 mol

when desired pH = 5.76

pH = pKa + log([acetate]/[acid])

⇒5.76 = 4.76 + log(x/(0.8-x))

⇒log(x/(0.8-x)) = 1

⇒x/(0.8-x) = 10

⇒x = 8 - 10x

⇒x = 8/11

⇒x= 0.73

[acetate] = x= 0.73

[acid] = 0.8-x = 0.07 M

number of mol = concentration * (volume in mL)

number of mol acetate to be added = 0.73*0.5 = 0.365 mol

number of mol acid to be added = 0.07*0.5 = 0.035 mol

Problem based on acetic acid required to maintain a certain pH

brainly.com/question/9240031

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