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Makovka662 [10]
2 years ago
11

Drag each positive ion to bond it with a negative ion to form the neutral ionic compound indicated.

Chemistry
1 answer:
weeeeeb [17]2 years ago
8 0

Answer:

1. NaCl

2. NH4F

3. MgO

4.LiCl

5. KI

6. CaO

Explanation:

In that order

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What is the answer if u know please answer
forsale [732]
An atom having 52 protons and 54 electrons would have an atomic number of 52 and a net charge of -2. This element would be 52 Te 2-, or choice A.
7 0
4 years ago
Describe how you would prepare 350 ml of 0.100 m c12h22o11 starting with 3.00l of 1.50 m c12h22o11
Leona [35]

To prepare 350 mL of 0.100 M solution from a 1.50 M solution, we simply have to use the formula:

M1 V1 = M2 V2

So from the formula, we will know how much volume of the 1.50 M we actually need.

 

1.50 M * V1 = 0.100 M * 350 mL

V1 = 23.33 mL

 

So we need 23.33 mL of the 1.50 M solution. We dilute it with water to a volume of 350 mL. So water needed is:

350 mL – 23.33 mL = 326.67 mL water

 

 

Steps:

1. Take 23.33 mL of 1.50 M solution

<span>2. Add 326.67 mL of water to make 350 mL of 0.100 M solution</span>

7 0
3 years ago
A certain weak acid, ha, has a ka value of 6.0×10−7. calculate the percent ionization of ha in a 0.10 m solution.
joja [24]
For the purpose, we will use the equation for determining the dissociation constant from concentration and <span>percent of ionization:

Kd = c </span>× α²

α = √(Kd/c) × 100%

Kd = 6.0×10⁻⁷

c(HA) = 0.1M

α = √(6.0×10⁻⁷/0.1)  × 100% =  0.23%

So, in the solution, the acid <span>percent of ionization will be just 0.23%.</span>

5 0
3 years ago
A 35.0-ml sample of 0.20 m lioh is titrated with 0.25 m hcl. What is the ph of the solution after 23.0 ml of hcl have been added
kiruha [24]
<h3><u>Answer;</u></h3>

pH = 12.33

<h3><u>Explanation;</u></h3>

The equation of reaction is :

LiOH(aq) + HCl(aq) --> LiCl(aq) + H2O(l)

Reactants left after the titrant is added;

Total Moles LiOH;

= 0.035L LiOH × (0.2moles/L)

= 0.007moles of LiOH

Moles of HCl;

= 0.023L HCl × (0.25moles/L)

= 0.00575moles HCl is the limiting reagent

Reacting amount of moles of LiOH;

= 0.0575 moles HCl *(1mole LiOH/1moles HCl)

=0.00575 moles LiOH (reacted)

Moles of LiOH left;

= 0.007moles total - 0.00575moles that react

= .00125 moles of LiOH (left)

LiOH is a strong base, which means that it ionizes completely.  

0.00125moles LiOH *(moles/0.058L) = 0.02155M of LiOH

LiOH(aq) --> Li+(aq) + OH-(aq)

[LiOH] = [OH-] = 0.02155 M

pOH = -log[OH-]

pOH = -log(0.02155)

pOH= 1.67

pH = 14 - pOH

pH = 14 - 1.67

pH = 12.33

7 0
3 years ago
Help would be much appreciated, I'm being timed<br> I'll mark brainliest !!
Vitek1552 [10]
Group 17

Please brainliesttt
7 0
3 years ago
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