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2 years ago

Help Me???????????????

Chemistry

1 answer:

ikadub [295]2 years ago

3 0

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Which of the following is the correct formula for Ca2+ reacting with NO2??

mariarad [96]

Answer:

B. Ca(NO2)2

Explanation:

Ions (charged atoms) combine with one another to form stable ionic compounds. In this case, calicium ion (Ca2+) is said to react with NO2, which has a charge of -1 i.e. +1-1(2) = 1-2 = -1.

This means that calcium has a charge of +2 while nitrite ion has charge of -1, hence, when they combine, they exchange their charges, which become their subscript as follows:

Ca2+ + NO2- → Ca(NO2)2

Ca(NO2)2 is a stable ionic compound called calcium nitrite. Notice that it takes two atoms of NO2- to react with one atom of Ca2+.

5 0

3 years ago

How many moles of lithium are in 18.2 grams of lithium

Doss [256]

Answer:

I think 2.6 sorry if its wrong

6 0

3 years ago

Read 2 more answers

(ii) Electrons are supplied to the external circuit. How and where is this done?<br> [2]

pentagon [3]

Answer:

The electrons are supplied by the species getting oxidized. They move from anode to the cathode in the external circuit. The external battery supplies the electrons. They enter through the cathode and come out through the anode

7 0

3 years ago

Nitrogen dioxide is produced by combustion in an automobile engine. For the following reaction, 0.377 moles of nitrogen monoxide

Contact [7]

Answer:

The amount of NO₂ that can be produced 8.533 g

Explanation:

According to question

2 NO(g) + O₂(g) → 2 NO₂(g)

Given

Moles of nitrogen monoxide = 0.377

Moles of oxygen = 0.278

$'For NO'=\frac{Mole}{Stoichiometry}=\frac{0.377}{2} =0.1855\\'For O_{2} '=\frac{0.278}{1}= 0.278\\$

Since 'NO' is the limiting reagent according to this ratio.

According to equation

2 moles NO reacts to form 2 moles NO₂

So, 0.1855 moles NO give = 0.1855 moles of NO₂

Mass of 1 mole NO₂ = 46 g/mole

Mass of 0.1855 moles = 46 x 0.1855 = 8.533 g

5 0

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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4 0

1 year ago