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azamat
3 years ago
15

2 Points

Chemistry
2 answers:
KiRa [710]3 years ago
8 0
The answer is C

A solution that cannot dissolve any more solute.
levacccp [35]3 years ago
4 0
B aaaaaaaaaaassaaaaaaaaaaaaaaaaaaaaaaaaaaaaAaaaaaaaaaaaa
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If the sample contained 2.0 moles of KClO3 at a temperature of 214.0 °C, determine the mass of the oxygen gas produced in grams
Westkost [7]

Answer : The mass of the oxygen gas produced in grams and the pressure exerted by the gas against the container walls is, 96 grams and 1.78 atm respectively.

Explanation : Given,

Moles of KCl_3 = 2.0 moles

Molar mass of O_2 = 32 g/mole

Now we have to calculate the moles of MgO

The balanced chemical reaction is,

2KClO_3\rightarrow 2KCl+3O_2

From the balanced reaction we conclude that

As, 2 mole of KClO_3 react to give 3 mole of O_2

So, 2.0 moles of KClO_3 react to give \frac{2.0}{2}\times 3=3.0 moles of O_2

Now we have to calculate the mass of O_2

\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2

\text{ Mass of }O_2=(3.0moles)\times (32g/mole)=96g

Therefore, the mass of oxygen gas produced is, 96 grams.

Now we have to determine the pressure exerted by the gas against the container walls.

Using ideal gas equation:

PV=nRT\\\\PV=\frac{w}{M}RT\\\\P=\frac{w}{V}\times \frac{RT}{M}\\\\P=\rho\times \frac{RT}{M}

where,

P = pressure of oxygen gas = ?

V = volume of oxygen gas

T = temperature of oxygen gas = 214.0^oC=273+214.0=487K

R = gas constant = 0.0821 L.atm/mole.K

w = mass of oxygen gas

\rho = density of oxygen gas = 1.429 g/L

M = molar mass of oxygen gas = 32 g/mole

Now put all the given values in the ideal gas equation, we get:

P=1.429g/L\times \frac{(0.0821L.atm/mole.K)\times (487K)}{32g/mol}

P=1.78atm

Thus, the pressure exerted by the gas against the container walls is, 1.78 atm.

7 0
3 years ago
How many gram-atoms are in 120-gram sample of calcium metal? How many atoms is this?
Elden [556K]
(sample g/1) X (1 mole/40.078(MW of Ca)) = moles of sample (moles of sample)(6.022 x 10^23( no of atoms)/ 1 mole) = # of atoms in a 120 g sample of calcium Avogadro's number=6.022x 10^23 atoms in 1 mole
3 0
3 years ago
Chloroform (CHCl3), an important solvent, is produced by a reaction between methane and chlorine. CH4(g) + 3 Cl2(g) CHCl3(g) + 3
PIT_PIT [208]
CH₄(g) + 3 Cl₂(g) → CHCl₃(g) + 3 HCl(g)
From the equation we notice that 1 mole of methane produces 1 mole of chloroform:
16 g Methane → 119.38 g Chloroform
?   g Methane → 37.5 g Chloroform
by cross multiplication:
= (16 * 37.5) / 119.38 = 5.0 g methane
4 0
3 years ago
How many electrons are there in carbonate ion, CO3^-2?
NISA [10]
It should be 24 electrons
3 0
3 years ago
. How many grams of magnesium chloride can be produced by reacting 2 moles of chlorine gas with excess magnesium bromide? ____Cl
dlinn [17]

Answer: 190 g of magnesium chloride can be produced by reacting 2 moles of chlorine gas with excess magnesium bromide.

Explanation:

The balanced chemical reaction is;

Cl_2+MgBr_2\rightarrow MgCl_2+Br_2

Cl_2 is the limiting reagent as it limits the formation of product and MgBr_2 is the excess reagent.

According to stoichiometry :

1 mole of Cl_2 produces = 1 mole of MgCl_2

Thus 2 moles of Cl_2 will produce=\frac{1}{1}\times 2=2moles  of MgCl_2

 Mass of MgCl_2=moles\times {\text {Molar mass}}=2moles\times 95g/mol=190g

Thus 190 g of magnesium chloride can be produced by reacting 2 moles of chlorine gas with excess magnesium bromide

5 0
2 years ago
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