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2 years ago

If the compound contains 35.06 % cl by mass, what is the identity of the metal?

Chemistry

1 answer:

Setler [38]2 years ago

4 0

Answer is: identity of the metal is gold (Au).

ω(Cl) = 35.06% ÷ 100%.

ω(Cl) = 0.3506; mass percentage of chlorine.

If we take 100 grams of the compound:

m(Cl) = ω(Cl) · m(compound).

ω(Cl) = 0.3506 · 100 g.

ω(Cl) = 35.06 g.

n(Cl) = m(Cl) ÷ M(Cl).

n(Cl) = 35.06 g ÷ 35.45 g/mol.

n(Cl) = 0.99 mol; amount of substance.

In molecule MCl₃: n(M) : n(Cl) = 1 : 3.

n(M) = 0.33 mol; amount of unknown metal.

M(M) = m(M) ÷ n(M).

M(M) = (100 g - 35.06 g) ÷ 0.33 mol.

M(M) = 196.8 g/mol; molar mass of the gold.

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Calculate the density of a material that has a mass of 50.5g and a volume of 14.5cm3.

White raven [17]

Density = mass / volume = 50.5/14.5 = 3.48 $g/cm ^{3}$ .

4 0

2 years ago

Mixtures of benzene and cyclohexane exhibit ideal behavior. A solution was created containing 1.5 moles of liquid benzene and 2.

goblinko [34]

Answer:

Vapour pressure of cyclohexane at 50°C is 490torr

Vapour pressure of benzene at 50°C is 90torr

Explanation:

Using Raoult's law, pressure of a solution is defined by the sum of the product sbetween mole fraction of both solvents and pressure of pure solvents.

$P_{solution} = X_{A}P^0_{A}+X_{B}P^0_{B}$

In the first solution:

$X_{cyclohexane}=\frac{2.5mol}{2.5mol+1.5mol} =0.625$

$X_{benzene}=\frac{1.5mol}{2.5mol+1.5mol} =0.375$

$340torr = 0.625P^0_{A}+0.375P^0_{B}$ (1)

For the second equation:

$X_{cyclohexane}=\frac{3.5mol}{3.5mol+1.5mol} =0.700$

$X_{benzene}=\frac{1.5mol}{3.5mol+1.5mol} =0.300$

$370torr = 0.700P^0_{A}+0.300P^0_{B}$ (2)

Replacing (2) in (1):

$340torr = 0.625P^0_{A}+0.375(1233.3-2.333P^0_{A})$

$340torr = 0.625P^0_{A}+462.5-0.875P^0_{A}$

-122.5torr = -0.250P°A

$P^0_{A} = 490 torr$

Vapour pressure of cyclohexane at 50°C is 490torr

And for benzene:

$370torr = 0.700*490torr+0.300P^0_{B}$

$P^0_{B}=90torr$

Vapour pressure of benzene at 50°C is 90torr

3 0

2 years ago

How many moles of oxygen atoms are present in 5 moles of Mg3(PO4)2

Irina18 [472]

How many moles of oxygen atoms are present in 5 moles of Mg3(PO4)2

All you have to do is to create a ratio between the molecule and the oxygen atom.

5 moles of Mg3(PO4)2 (4x2 moles O/1 mole Mg3(PO4)2) = 40 moles of oxygen

5 0

3 years ago

A 3.8-mol sample of KClO3 was decomposed according to the equation. How many moles of O2 are formed assuming 100% yield?

kari74 [83]

Answer:

5.7 moles of O2

Explanation:

We'll begin by writing the balanced decomposition equation for the reaction. This is illustrated below:

2KClO3 —> 2KCl + 3O2

From the balanced equation above,

2 moles of KClO3 decomposed to produce 3 moles of O2.

Next, we shall determine the number of mole of O2 produced by the reaction of 3.8 moles of KClO3.

Since 100% yield of O2 is obtained, it means that both the actual yield and theoretical yield of O2 are the same. Thus, we can obtain the number of mole of O2 produced as follow:

From the balanced equation above,

2 moles of KClO3 decomposed to produce 3 moles of O2.

Therefore, 3.8 moles of KClO3 will decompose to produce = (3.8 × 3)/2 = 5.7 moles of O2.

Thus, 5.7 moles of O2 were obtained from the reaction.

3 0

2 years ago

What mass of Hg will occupy a volume of 75.0 mL?

k0ka [10]

75.0 mL in liters:

75.0 / 1000 => 0.075 L

1 mole -------------------- 22.4 L ( at STP)
( moles Hg) ------------- 0.075 L

moles Hg = 0.075 x 1 / 22.4

moles = 0.075 / 22.4

= 0.00334 moles of Hg

Hg => 200.59 u

1 mole Hg ----------------- 200.59 g
0.00334 moles Hg ----- ( mass Hg )

mass Hg = 200.59 x 0.00334 / 1

mass Hg = 0.6699 / 1

= 0.6699 g of Hg

7 0

2 years ago