→ 
Explanation:
- The products formed are chromic chloride and cobalt.
Chromium + Cobaltous Chloride = Chromic Chloride + Cobalt
- Type of reaction is Single Displacement (Substitution) which is there is a displacement of one atom.
Reactants used in the reaction are -
- Chromium

- Cobaltous Chloride

Products formed in the reaction are -
- Chromic Chloride

- Cobalt

Hence, the chemical reaction is as follows -
→
For balancing the above chemical equation we need to add a coefficient of 2 in front of chromium and of 3 in front of cobalt(II)chloride on right-hand-side while of 2 in front of chromium chloride and of 3 in front of carbon monoxide on left-hand-side of the equation.
Hence, the balanced equation is -
→ 
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A, it says on here good luckkkkkk
Answer:
16.89g of PbBr2
Explanation:
First, let us calculate the number of mole of Pb(NO3)2. This is illustrated below:
Molarity of Pb(NO3)2 = 0.595M
Volume = 77mL = 77/1000 = 0.077L
Mole =?
Molarity = mole/Volume
Mole = Molarity x Volume
Mole of Pb(NO3)2 = 0.595x0.077
Mole of Pb(NO3)2 = 0.046mol
Convert 0.046mol of Pb(NO3)2 to grams as shown below:
Molar Mass of Pb(NO3)2 =
207 + 2[ 14 + (16x3)]
= 207 + 2[14 + 48]
= 207 + 2[62] = 207 +124 = 331g/mol
Mass of Pb(NO3)2 = number of mole x molar Mass = 0.046 x 331 = 15.23g
Molar Mass of PbBr2 = 207 + (2x80) = 207 + 160 = 367g/mol
Equation for the reaction is given below:
Pb(NO3)2 + CuBr2 —> PbBr2 + Cu(NO3)2
From the equation above,
331g of Pb(NO3)2 precipitated 367g of PbBr2
Therefore, 15.23g of Pb(NO3)2 will precipitate = (15.23x367)/331 = 16.89g of PbBr2