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3 years ago

6

How many valence electrons are in an atom of phosphorus?

1 answer:

Helen [10]3 years ago

8 0

How many valence electrons are in an atom of phosphorus?

a. 2
c. 4
b. 3
d. 5 Please select the best answer from the choices provided A B C D

There are five valence electrons are in an atom of phosphorus. The answer is letter D.

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Each positive is blanched with a negative. true or false

dimulka [17.4K]

Answer:

True is the answer

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2 years ago

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2 years ago

In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].

Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose + $H^+\rightarrow$ fructose+ glucose

The rate law of the reaction is given as:

$R=k[H^+][sucrose]$

$[H^+]=0.01M$

[sucrose]= 1.0 M

$R=k[0.01M][1.0 M]$ ..[1]

a)

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

$R'=[0.01 M][2.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

$R'=[0.01 M][0.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)

The rate of the reaction when $[H^+]$ is changed to 0.001 M = R'

$R'=[0.0001 M][1.0 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}$

$R'=0.01\times R$

If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d)

The rate of the reaction when [sucrose] and $[H^+]$ both are changed to 0.1 M = R'

$R'=[0.1M][0.1M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}$

$R'=1\times R$

If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0

3 years ago

Any atom is most stable when it’s outermost energy level contains

slamgirl [31]

Explanation:

Atoms are at their most stable when their outermost energy level is either empty of electrons or filled with electrons. Sodium atoms have 11 electrons. Two of these are in the lowest energy level, eight are in the second energy level and then one electron is in the third energy level.

6 0

2 years ago

In a single displacement reaction between Sodium Phosphate and Barium, how much of each product (in grams) will be formed from 1

mixer [17]

Answer:

14.6 g of barium phosphate

3.35 g of sodium metal

Explanation:

2Na3PO4(aq) + 3Ba(s) -------> Ba3(PO4)2(aq) + 6Na(s)

The first step in any such reaction is to but down the balanced reaction equation according to the stoichiometry of the reaction.

The two products formed are barium phosphate and sodium metal.

Number of moles of barium corresponding to 10.0g of barium = mass of barium/ molar mass of barium

Molar mass of barium = 137.327 g

Number of moles of barium = 10/137.327

Number of moles of barium = 0.0728 moles

For barium phosphate;

3 moles of barium yields 1 mole of barium phosphate

0.0728 moles yields 0.0728 moles × 1/3 = 0.0243 moles of barium phosphate

Molar mass of barium phosphate = 601.93 g/mol

Therefore mass of barium phosphate = 0.0243 moles × 601.93 g/mol = 14.6 g of barium phosphate

For sodium metal

3 moles of barium yields 6 moles of sodium metal

0.0728 moles of barium yields 0.0728 × 6 / 3 = 0.1456 moles of sodium

Molar mass of sodium metal= 23 gmol-1

Mass of sodium metal= 0.1456g × 23 gmol-1 = 3.35 g of sodium metal

4 0

3 years ago