What is what is the balanced formula equation for Solutions of lead nitrate and lithium chloride

What volume would be occupied by 100. g of oxygen gas at a pressure of 1.50 atm and a temperature of

PSYCHO15rus [73]

Answer:

Use PV = nRT. Don't forget T must be in kelvin.

Explanation:

8 0

3 years ago

What volume of each solution contains 0.14 mol of KCl? Express your answer using two significant figures.

trasher [3.6K]

Answer:

Solution given:

1 mole of KCl $\rightarrow$ 22.4l

1 mole of KCl $\rightarrow$ 74.55g

we have

0.14 mole of KCl $\rightarrow$ 74.55*0.14=10.347g

74.55g of KCl $\rightarrow$ 22.4l

10.347 g of KCl $\rightarrow$ 22.4/74.55*10.347=3.11litre

volume of each solution contains 0.14 mol of KCl contain 3.11litre.

5 0

3 years ago

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Calculate the mass of oxalic acid dihydrate needed to prepare 500.0 ml

VARVARA [1.3K]

Moles= molarity x liter
=0.2500 M x 0.500 L
= 0.1250 mol

mass C2H2O4 . 2H2O = 0.1250 mol x (126.068 g / 1 mol)
=15.76 g

6 0

3 years ago

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How many moles of NaOH is needed to neutralize 45.0ml of 0.30M H2SeO4?

kiruha [24]

Answer:

0.027 mole of NaOH.

Explanation:

We'll begin by obtaining the number of mole H2SeO4 in 45mL of 0.30M H2SeO4

This is illustrated below:

Molarity of H2SeO4 = 0.3M

Volume of solution = 45mL = 45/1000 = 0.045L

Mole of H2SeO4 =...?

Mole = Molarity x Volume

Mole of H2SeO4 = 0.3 x 0.045

Mole of H2SeO4 = 0.0135 mole

Next, the balanced equation for the reaction. This is given below:

H2SeO4 + 2NaOH –> Na2SeO4 + 2H2O

From the balanced equation above,

1 mole of H2SeO4 required 2 moles of NaOH.

Therefore, 0.0135 mole of H2SeO4 will require = 0.0135 x 2 = 0.027 mole of NaOH.

Therefore, 0.027 mole of NaOH is needed for the reaction.

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3 years ago

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Gallium oxide, Ga.Oy, forms when gallium is combined with oxygen. A 1.25 g of Ga is allowed to react with excess oxygen and 1.68

mrs_skeptik [129]

Answer : The chemical formula of a compound is, $Ga_2O_3$

Solution : Given,

Mass of gallium = 1.25 g

Mass of gallium oxide = 1.68 g

Mass of oxygen = Mass of gallium oxide - Mass of gallium

Mass of oxygen = 1.68 - 1.25 = 0.43 g

Molar mass of Ga = 69.72 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of Ga = $\frac{\text{ given mass of Ga}}{\text{ molar mass of Ga}}= \frac{1.25g}{69.72g/mole}=0.0179moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{0.43g}{16g/mole}=0.027moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Ga = $\frac{0.0179}{0.0179}=1$

For O = $\frac{0.027}{0.0179}=1.5$

The ratio of Ga : O = 1 : 1.5

To make in whole number we multiple ratio by 2, we get:

The ratio of Ga : O = 2 : 3

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $Ga_2O_3$

The empirical formula weight = 2(69.72) + 3(16) = 187.44 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{187.44}{187.44}=1$

Molecular formula = $(Ga_2O_3)_n=(Ga_2O_3)_1=Ga_2O_3$

Therefore, the chemical of the compound is, $Ga_2O_3$

5 0

3 years ago