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zaharov [31]
3 years ago
10

If 100 ml of some pb(no3)2 solution is mixed with 100 ml of 6.50 x 10−2 m nacl solution, what is the maximum concentration of th

e pb(no3)2 solution added if no solid pbcl2 forms? (assume ksp = 2.00 x 10−5 m at this temperature.) enter the concentration in m.
Chemistry
1 answer:
Alex73 [517]3 years ago
7 0
Suppose X (in unit M) be the required maximum concentration of the Pb(NO₃)₂ solution added after mixing with 100 ml of 6.5 x 10⁻² M NaCl, the Pb²⁺ concentration is:
[Pb²⁺] = (X*100) / 200 = 0.5 X
The concentration of Cl⁻ is:
[Cl⁻] = (100 / 200) * (6.5 x 10⁻² M) = 0.0325 M
So:
Ksp = 2.00 x 10⁻⁵ = [Pb²⁺] [Cl⁻]² = (0.5X) * (0.0325)²
X = (2.00 x 10⁻⁵) / (5.28 x 10⁻⁴) = 0.038 M
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If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is Group of answer choices
BaLLatris [955]

Answer:

The water is completely vaporized at this stage.

Explanation:

The complete question is

If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is

-boiling

-completely vaporized

-frozen solid

-decomposed

-still a liquid

Energy added = 50 kJ = 50000 J

mass of water = 15.5 g = 0.0155 kg

temperature of water = 10 °C

We know that the energy posses by a mass of water at a given temperature is given as

H = mcT

where H is the energy possessed by the mass of water

m is the mass of the water

c is the specific heat capacity of water = 4200 J/ kg- °C

T is the temperature of the water

substituting values, the energy of this amount of water is

H = 0.0155 x 4200 x 10 = 651 J

If 50 kJ is added to the water, the energy increases to

50000 J + 651 J = 50651 J

Temperature of this water at this stage will be gotten from

H = mcT

we solve for the new temperature

50651 = 0.0155 x 4200 x T

50651 = 65.1 x T

T = 50651/65.1 = 778.05 °C

This temperature is well over 100 °C, which is the vaporization temperature of water, but less than 3000 °C for its molecules to decompose.

3 0
3 years ago
How many moles of mercury are equivalent to 3.46 x 1023 atoms?
olga55 [171]

Answer:

3.02 X1023 atoms Ag limol. - - 0.50 1 moles. 6.02241023 atoms.

4 0
3 years ago
Determine the AMOUNT OF NO2, LIMITING REACTANT, AND THE AMOUNT AND NAME OF EXCESS REACTANT.
zepelin [54]

Answer:

balanced equation mole ratio 5 2 mol NO/1 mol O2

10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2

20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO

actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2

Because the actual mole ratio of NO:O2 is larger than the balanced equation mole

ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.

Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO

0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO

Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2

Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N

Explanation:

3 0
3 years ago
A. Calculate the empirical formula of a molecule with percent compositions: 55.3% potassium (K), 14.6% phosphorus (P), and 30.1%
Otrada [13]
The way you calculate the empirical formula is to firstly assume 100g. To find each elements moles you take each elements percentage listed, times it by one mole and divide it by its atomic mass. (ex: moles of K =55.3g x 1 mole/39.1g, therefore there is 1.41432225 moles of Potassium) Once you’ve completed this for every element you list each elements symbol beside it’s number of moles and divide by the smallest number because it can only go into its self once. After you’ve done this, you’ve found your empirical formula, which is the simplest whole number ratio of atoms in a compound. I’ve added an example of a empirical question I completed last semester :)

6 0
3 years ago
Read 2 more answers
At standard ambient temperature and pressure (SATP), a gas has density of 1.5328g/L. What is the molar mass of the gas?
ahrayia [7]

The standard ambient temperature and pressure are

Temperature =298 K

Pressure = 1atm

The density of gas is 1.5328 g/L

density = mass of gas per unit volume

the ideal gas equation is

PV = nRT

P = pressure = 1 atm

V = volume

n = moles

R= gas constant = 0.0821 Latm/mol K

T = 298 K

moles = mass / molar mass

so we can write

n/V = density / molar mass

Putting values

Pressure=\frac{nRT}{V}=\frac{massXRT}{VXmolarmass}

Pressure=\frac{densityXRT}{molarmass}

molarmass=\frac{densityXRT}{Pressure}=\frac{1.5328X0.0821X298}{1}=37.50

Thus molar mass of gas is 37.50g/mol

6 0
3 years ago
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