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3 years ago

How do you detemine the charge of an element based on jts valence electrons

1 answer:

8 0

Look at the protons and electrons and rememeber metals want to get rid of electrons to be perfect. Nonmetals are wanting to take in. Metals are negatives in electrons. Opposite for nonmetals.

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Robert has pure samples of both D-ribose and D-arabinose, but he forgot to label them. He only has some nitric acid, the reagent

Sauron [17]

Answer:

Following are the responses to the given question:

Explanation:

Since HN03 is an oxidation substance D-ribose u.ith oxidized to form in rubric acid Ribose is chiral, but rubric acid is achiral because of its symmetry mirror level, Hence no infrared roster in the sample holder is observed.

Please find the attached file.

D-Arabinose, on either hand, gives optical aldaric acid with such a net optical rotation observed inside the polarimeter for diagnosis with HN03.

4 0

3 years ago

Irrigation is a method used to water farm land. Identify two effects irrigation has on the water cycle and environment. A Irriga

sergiy2304 [10]

Answer:

A and C

Explanation:

6 0

3 years ago

If 4.12 l of a 0.850 m-h3po4 solution is be diluted to a volume of 10.00 l, what is the concentration of the resulting solution?

statuscvo [17]

The process in which the concentration of the solution is lessened by the addition of water is said to be dilution and equation of dilution relates the initial concentration and volume of stock solution with the final concentration and volume of the solution.

Formula is given by:

$C_{1}V_{1}=C_{2}V_{2}$ (1)

where,

$C_{1}$ is the initial concentration

$V_{1}$ is the initial volume

$C_{2}$ is the final concentration

$V_{2}$ is the final volume

Now,

$C_{1}$ = 0.850 M

$V_{1}$ = 4.12 L

$C_{2}$ =?

$V_{2}$ = 10.00 L

Substitute the give values in formula (1),

$0.850 M\times 4.12L=C_{2}\times 10.00 L$

$C_{2} =\frac{0.850 M\times 4.12L}{10.00 L}$

= $0.3502 M$

Thus, the final concentration of the $H_{3}PO_{4}$ solution = $0.3502 M$

5 0

3 years ago

Some nitrogen is held in a 2.00-L tank at a pressure of 3.00 atm. The tank is connected to a 5.00-L tank that is completely empt

leva [86]

Answer:

0.857 atm

Explanation:

The nitrogen stops owing when it fulfills both of the tanks (the gas molecules intend to fulfill all the space they are). So the tanks will have the same pressure, and the final volume will be the volume of the two tanks.

For Boyle's law:

P1*V1 = P2*V2

Where P1 is the initial pressure (3.00 atm), V1 is the initial volume(2.00 L), P2 is the final pressure, and V2 is the final volume (2.00 + 5.00 = 7.00 L).

3.00*2.00 = P2*7.00

7.00P2 = 6.00

P2 = 0.857 atm

7 0

3 years ago

Consider the reaction of 2.5 grams of Li (s) reacting with 0.5 grams of N2 (g) to produce Li3N (s). A) How many total grams of L

vaieri [72.5K]

Answer:

A) The amount in grams of Li₃N produced is 1.243 g

B) N₂, is the limiting reagent

The mass of the non-limiting reagent, Li, remaining after the reaction is completed is 1.757 g

Explanation:

The given parameters are;

The mass of Li(s) = 2.5 grams

The mass of N₂ (g) = 0.5 grams

The chemical equation for the reaction can be presented as follows;

6 Li (s) + N₂ (g) → 2 Li₃N

Therefore, 6 moles of Li reacts with 1 mole of N₂ to produce 2 moles of Li₃N

The molar mass of Li = 6.941 g/mol

The molar mass of N₂ = 28.0134 g/mol

The number of moles of a reactant or product, n is given by the relation;

n = Mass of substance/(Molar mass of the substance)

For lithium, Li, n = 2.5/6.941 = 0.3602 moles

For Nitrogen gas, N₂, n = 0.5/28.0134 = 0.01785 moles

A) Given that 1 mole of N₂ to produces 2 moles of Li₃N

0.01785 moles of N₂ will produces 2×0.01785 = 0.0357 moles of Li₃N

The molar mass of Li₃N = 34.83 g/mol

The mass of Li₃N = 34.83 g/mol × 0.0357 moles = 1.243 g

B) 6 moles of Li reacts with 1 mole of N₂ to produce 2 moles of Li₃N

0.3602 moles will reacts with 1/6×0.3602 = 0.06003 mole of N₂

Therefore, N₂, is the limiting reagent and we have;

0.01785 moles of N₂ will react with 6×0.01785 = 0.1071 moles of Li

The number of of moles of Li left = 0.3602 - 0.1071 =0.2531 moles

The mass of lithium left = 0.2531 moles × 6.941 g/mol = 1.757 g

The mass of lithium remaining after the reaction is completed = 1.757 g.

4 0

3 years ago