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aliina [53]
3 years ago
14

Write the net ionic equation for the reaction of sodium chloride with magnesium hydroxide. Include the complete ionic equation a

nd label spectator ions. Also be sure to label the reactants and products as solid, liquid, gas, or aqueous solutions.
Chemistry
2 answers:
Lera25 [3.4K]3 years ago
8 0
Idk idk idk idk Idk idk idk idk idk Idk idk idk idk idk Idk idk idk idk idk
svp [43]3 years ago
7 0
This a long process...
first you write down the equation,in this case:
2NaCl (aq) + Mg(OH)2 (s) ———> MgCl2 (aq) + 2NaOH (aq) .
next you break them up into ions :
(only break up the aqueous solutions)
(2Na+) + (2Cl-) + Mg(OH)2 {leave it together} ———> (Mg2+) + (2Cl-) + (2Na+) + (2OH-).
then look at both sides of equation and cancel the ions that appear twice .
In this case: Cl and Na.
now the remaining is your final net atomic equation:
Mg(OH)2———-> Mg2 + 2OH.
You might be interested in
If the density of nitrogen gas at a certain pressure and temperature is 1.20 g/l, how many moles of nitrogen gas are in 15.0 l
Daniel [21]
Density of nitrogen= 1.20 g/l
no. of grams of nitrogen=m= (1.20g/l)(15.0l)= 18 g
molar mass of nitrogen=M= (14*2)= 28 g/mol

no. of moles of nitrogen=m/M=18/28= 0.643 moles

8 0
3 years ago
A sample of damp air in a 1.00 L container exerts a total pressure of 741.0 torr at 20 oC; but when it is cooled to -10 oC, the
AlexFokin [52]

From the information presented in the question, the number of molecules present of water present is obtained 2.41 × 10^21 molecules.

From the information we have;

Volume of the damp air =  1 L

Pressure of the damp air =  741.0 torr or 0.975 atm

Temperature of the gas = 20 oC + 273 = 293 K

R = 0.082 atm LK-1mol-1

Number of moles = ?

n =PV/RT

n = 0.975 × 1/0.082 ×  293

n = 0.041 moles

Volume of water vapor = 1 L

Temperature of water = -10 oC + 273 = 263 K

Pressure of the gas = 607.1 torr or 0.799 atm

R = 0.082 atm LK-1mol-1

n= PV/RT

n = 0.799 × 1/ 0.082 × 263

n = 0.037 moles

Number of moles of water = 0.041 moles -  0.037 moles = 0.004 moles

If 1 mole = 6.02 × 10^23 molecules

0.004 moles = 0.004 moles × 6.02 × 10^23 molecules/1 mole

= 2.41 × 10^21 molecules

Learn more: brainly.com/question/2510654

5 0
2 years ago
Convert 779 gallons to cubic yards
AfilCa [17]
The answer would be About 3.857yd3
7 0
3 years ago
I need to check some chemistry questions. Help with any question is appreciated! :) Please include an explanation with your conc
finlep [7]

1. 4.67 kg; 2. 4.8 ×10^5 kg; 3. 0.106 cm^3; 4. 1.7 g/cm^3

<em>Q1. Mass of Hg </em>

Mass = 345 mL × (13.53 g/1 mL) = 4670 g = 4.67 kg

<em>Q2. Mass of Pb </em>

<em>Step 1</em>. Calculate the <em>volume of the Pb</em>.

<em>V = lwh</em> = 6.0 m × 3.5 m × 2.0 m = 42.0 m^3

<em>Step 2</em>. Calculate the <em>mass of the Pb</em>.

Mass = 42.0 m^3 × (11 340 kg/1 m^3) = 4.8 × 10^5 kg

<em>Q3. Volume of displaced water </em>

Volume of Ag = 0.987 g × (1 cm^3/9.320 g) = 0.106 cm^3

<em>Archimedes</em>: volume of displaced water = volume of Ag = <em>0.106 cm^3</em>

<em>4. Density of metal </em>

<em>Step 1</em>. Convert <em>ounces to grams </em>

Mass = 3.35 oz × (28.35 g/1 oz) = 94.97 g

<em>Step 2</em>. Calculate the <em>volume in cubic inches </em>

<em>V = lwh</em> = 3.0 in × 2.5 in × 0.45 in = 3.38 in^3

<em>Step 3</em>. Convert <em>cubic inches to cubic centimetres</em><em> </em>

<em>V</em> = 3.38 in^3 × (2.54 cm/1 in)^3 = 55.3 cm^3

<em>Step 4</em>. Calculate the <em>density</em>

ρ = <em>m</em>/<em>V</em> = (94.97 g/55.3 cm^3) = 1.7 g/cm^3 (magnesium?)

4 0
3 years ago
A compound is found to contain 42.88 % carbon and 57.12 % oxygen by weight. To answer the questions, enter the elements in the o
Ipatiy [6.2K]

Answer:

The empirical formule is CO

Explanation:

Step 1: Data given

Suppose the mass of a compound is 100 grams

Suppose the compound contains:

42.88 % C = 42.88 grams C

57.12 % O = 57.12 grams O

Molar mass C = 12.01 g/mol

Molar mass O = 16.0 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles C = mass C / molar mass C

Moles C = 42.88 grams / 12.01 g/mol

Moles C = 3.57 moles

Moles O = 57.12 grams / 16.0 g/mol

Moles O = 3.57 moles

Step 3: Calculate the mol ratio

We divide by the smallest amount of moles

C: 3.57 moles / 3.57 moles = 1

O: 3.57 moles / 3.57 moles = 1

The empirical formule is CO

3 0
3 years ago
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