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2 years ago

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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Nina [5.8K]

First, let us find the corresponding amount of moles H₂ assuming ideal gas behavior.

PV = nRT
Solving for n,
n = PV/RT
n = (6.46 atm)(0.579 L)/(0.0821 L-atm/mol-K)(45 + 273 K)
n = 0.143 mol H₂

The stoichiometric calculations is as follows (MW for XeF₆ = 245.28 g/mol)
Mass XeF₆ = (0.143 mol H₂)(1 mol XeF₆/3 mol H₂)(245.28 g/mol) = <em>11.69 g</em>

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2 years ago

A mixture of N2, O2, and Ar has mole fractions of 0.25, 0.65, and 0.10, respectively. What is the pressure of N2 if the total pr

s344n2d4d5 [400]

Answer:

Partial pressure of nitrogen gas is 0.98 bar.

Explanation:

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gases.

$P=p_{N_2}+p_{O_2}+p_{Ar}$

$p_{N_2}=P\times \chi_{N_2}$

$p_{O_2}=P\times \chi_{O_2}$

$p_{Ar}=P\times \chi_{Ar}$

where,

$P$ = total pressure = 3.9 bar

$p_{N_2}$ = partial pressure of nitrogen gas

$p_{O_2}$ = partial pressure of oxygen gas

$p_{Ar}$ = partial pressure of argon gases

$\chi_{N_2}$ = Mole fraction of nitrogen gas = 0.25

$\chi_{O_2}$ = Mole fraction of oxygen gas = 0.65

$\chi_{Ar}$ = Mole fraction of argon gases = 0.10

Partial pressure of nitrogen gas :

$p_{N_2}=P\times \chi_{N_2}=3.9 bar\times 0.25 =0.98 bar$

Partial pressure of oxygen gas :

$p_{N_2}=P\times \chi_{O_2}=3.9 bar\times 0.65=2.54 bar$

Partial pressure of argon gas :

$p_{N_2}=P\times \chi_{Ar}=3.9 bar\times 0.10=0.39 bar$

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2 years ago

(f) what is the observed rotation of 100 ml of a solution that contains 0.01 mole of d and 0.005 mole of l? (assume a 1-dm path

never [62]

<span>Answer: .01 moles of D to .005 moles of L ~ so, .01+.005 = .015 total; using this total value, divide the portions of D and L. so .01/.015 to .005/.015 ~ 67% D to 33% L. And thus, the enantiomer excess will be 34%.</span>

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2 years ago

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For each pair below, select the sample that contains the largest number of moles. Pair A 2.50 g O2 2.50 g N2

raketka [301]

Answer:

Explanation:

Pair 2.50g of O₂ and 2.50g of N₂

The atoms sample with the largest number of moles since the masses are the same would be the one with lowest molar mass according the the equation below:

Number of moles = $\frac{mass }{molarmass}$

Atomic mass of O = 16g and N = 14g

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Number of moles of O₂ = $\frac{2.5}{32}$ = 0.078mole

Number of moles of N₂ = $\frac{2.5}{28}$ = 0.089mole

We see that N₂ has the largest number of moles

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2 years ago

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To prepare a 2 M solution of potassium nitrate (KNO3), which quantities must be measured? The mass of the and the volume of the

Olin [163]

Answer:

The mass of the solute and the volume of the solution.

Explanation:

Hello,

In this case, given the formula of molarity:

$M=\frac{n_{solute}}{V_{solution}}$

In such a way, since the moles could not be directly measured, we must measure the mass of the solute and by using its molar mass, one could compute its moles. Moreover, since the solution is composed by the solvent (typically water) and the solute, we consequently must measure the volume of the solution needed for the preparation of such concentration-known solution. In such a way, we can actually prepare the required solution.

Best regards.

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