The chemical formula for cesium oxide is Cs2O. What is the charge of cesium? +1 +2 –1 –2

Chemistry

2 answers:

Luda [366]2 years ago

8 0

Answer:

The correct answer, the charge of Cesium is +1

Explanation:

Hi!

Let's solve this!

The compound is: Cs2O.

We know that the oxygen valence number is -2.

We also know that the compound has to be balanced, so to balance the charges we need +2. As there are two Cs, we perform the following operation:

2 * (+ 1) -2 = + 2-2 = 0

We conclude that the correct answer, the charge of Cesium is +1

aalyn [17]2 years ago

5 0

The Charge Of Cesium Is +1

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A chemist reacted 57.50 grams of sodium metal with an excess amount of chlorine gas. The chemical reaction that occurred is show

Marizza181 [45]

The actual yield is 125.6
Balanced equation: 2Na + Cl2–>2NaCl

To find the theoretical yield :

57.50gNa ( 1mol/23gNa ) (2mol NaCl/2molNa)(58.44g NaCl / 1molNaCl) = 146.1g NaCl
Theoretical yield : 146.1g
Percentage yield: 86% or 0.86
(0.86)(146.1) =125.6

8 0

2 years ago

25 ml of 2.0 M silver acetate reacts with grams 35 ml of 1.0 M Calcium chloride

sp2606 [1]

Answer:

$m_{AgCl}=7.15gAgCl\\m_{Ca\ Acet}=3.15gCa\ Acet\\$

Explanation:

Hello,

In this case, with the given information, we can identify the limiting reactant and compute the theoretical yield for the undergoing chemical reaction:

$2CH_3COOAg+CaCl_2\rightarrow (CH_3COO)_2Ca+2AgCl$

Thus, with the given concentrations and volumes we compute the available moles of silver acetate:

$n_{Ag\ Acet}=2.0mol/L*0.025mL=0.05mol$

Then, the moles of silver acetate that are consumed by 35 mL of 1.0 M calcium chloride:

$n=0.035mL*1.0molCaCl_2/L*\frac{2mol Ag\ Acet}{1molCaCl_2} =0.07mol Ag\ Acet$

Therefore, since there are less available moles, it is the limiting reactant, for that reason, the theoretical yields of both calcium acetate and silver acetate are:

$m_{AgCl}=0.05mol Ag\ Acet*\frac{2molAgCl}{2mol Ag\ Acet} *\frac{143gAgCl}{1molAgCl} \\\\m_{AgCl}=7.15gAgCl\\\\m_{Ca\ Acet}=0.05mol Ag\ Acet*\frac{1molmol Ca\ Acet}{2molmol Ag\ Acet} *\frac{126gmol Ca\ Acet}{1mol Ca\ Acet} \\\\m_{Ca\ Acet}=3.15gCa\ Acet$