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svetlana [45]
3 years ago
10

Need the ionic formula

Chemistry
1 answer:
JulijaS [17]3 years ago
3 0

Answer:

To find the formula of an ionic compound, first identify the cation and write down its symbol and charge. Then, identify the anion and write down its symbol and charge. Finally, combine the two ions to form an electrically neutral compound.

Explanation:

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What is the formula name for Co3N2
Ede4ka [16]

Answer:

Cobaltous Nitride,I think so anyway.......

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2. What is the total mass of Chlorine in ZnCl,?
Vedmedyk [2.9K]

Answer:

35.5

Explanation:

35.5.

relative moleculer mass

6 0
2 years ago
A gas held at constant volume is heated from -5 degrees C to 50 degrees C, if the initial pressure is 1.0 atm, what is the new p
Hunter-Best [27]
1.21 atm, assuming the gas behaves ideally
8 0
3 years ago
45.7 grams of calcium chloride reacts with an excess of aluminum oxide. How many grams of aluminum chloride will be produced
damaskus [11]
Molar mass (CaCl2) = 40.1 +2*35.5 = 111.1 g/mol
Molar mass (AlCl3) = 27.0 +3*35.5= 133.5 g/ mol

                                               
3CaCl2+Al2O3 -------->3CaO +2AlCl3
mole from reaction              3 mol                                              2 mol
mass from reaction         3mol* 111.1g/mol                             2 mol*133.5g/mol
                                               333.3 g                                            267.0 g
mass from problem              45.7 g                                               x g

Proportion:
  333.3 g  CaCl2  -------   267.0 g AlCl3
  45.7 g   CaCl2   --------   x g    AlCl3

x=45.7*267.0/333.3= 36.6 g AlCl3
5 0
3 years ago
I am having some trouble figuring out how to approach the following problem: A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is ti
Tanya [424]
<span>We look at the end of the day:

n(HNO3) added = 0.500*17.0/1000 = 0.00850 mol
n(NH3) = 0.200*75.0/1000 - 0.00850 = 0.00650 mol
[NH3] left = 0.00650*1000/(17.0+75.0) = 0.070652
M [OH-] = Kb * [NH3] = 0.070652*1.8*10^(-5) = 1.27174 x 10^(-6)
pOH = -log[OH-] ≈ 5.8956 pH = 14 - pOH ≈ 8.10</span>
4 0
2 years ago
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