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2 years ago

Compare and contrast these isotopes: uranium -239, uranium -238, and uranium -235

Chemistry

2 answers:

kykrilka [37]2 years ago

4 0

They all have the same number of protons and electrons but differ in the number of neutrons because they are isotopes.

PSYCHO15rus [73]2 years ago

3 0

Answer: They have same number of protons but differ in number of neutrons.

Explanation:

Isotopes are defined as the species which have same number of neutrons but differ in the number of neutrons.

Uranium is the 92nd element of the periodic table.

Atomic number is defined as the number of protons that are present in an atom.

Atomic number = Number of protons

Mass number is defined as the sum of number of protons and neutrons that are present in an atom.

Mass number = Number of protons + Number of neutrons

For the given isotopes: $_{92}^{239}\textrm{U},_{92}^{238}\textrm{U}\text{ and }_{92}^{235}\textrm{U}$

For $_{92}^{239}\textrm{U}$

Number of protons = 92

Number of neutrons = 239 - 92 = 147

For $_{92}^{238}\textrm{U}$

Number of protons = 92

Number of neutrons = 238 - 92 = 146

For $_{92}^{235}\textrm{U}$

Number of protons = 92

Number of neutrons = 235 - 92 = 143

Hence, they have same number of protons but differ in number of neutrons.

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Which of the following statements correctly illustrates the difference between weather and climate? A Tonight’s cool, breezy wea

Ratling [72]

i actually dont know what is it?

Explanation:

5 0

2 years ago

When 13.6 g of calcium chloride, CaCl2, was dissolved in 100.0 mL of water in a coffee cup calorimeter, the temperature rose fro

DanielleElmas [232]

Answer:

THE ENTHALPY OF SOLUTION IS 3153.43 J/MOL OR 3.15 KJ/MOL.

Explanation:

1. write out the variables given:

Mass of Calcium chloride = 13.6 g

Change in temperature = 31.75°C - 25.00°C = 6.75 °C

Density of the solution = 1.000 g/mL

Volume = 100.0 mL = 100.0 mL

Specific heat of water = 4.184 J/g °C

Mass of the water = unknown

2. calculate the mass of waterinvolved:

We must first calculate the mass of water in the bomb calorimeter

Mass = density * volume

Mass = 1.000 * 100

Mass = 0.01 g

3. calculate the quantity of heat evolved:

Next is to calculate the quantity of heat evolved from the reaction

Heat = mass * specific heat of water * change in temperature

Heat = mass of water * specific heat *change in temperature

Heat = 13.6 g * 4.184 * 6.75

Heat = 13.6 g * 4.184 J/g °C * 6.75 °C

Heat = 384.09 J

Hence, 384.09J is the quantity of heat involved in the reaction of 13.6 g of calcium chloride in the calorimeter.

4. calculate the molar mass of CaCl2:

Next is to calculate the molar mas of CaCl2

Molar mass = ( 40 + 35.5 *2) = 111 g/mol

The number of moles of 13.6 g of CaCl2 is then:

Number of moles of CaCl2 = mass / molar mass

Number of moles = 13.6 g / 111 g/mol

Number of moles = 0.1225 mol

So 384.09 J of heat was involved in the reaction of 1.6 g of CaCl2 in a calorimter which translates to 0.1225 mol of CaCl2..

5. Calculate the enthalpy of solution in kJ/mol:

If 1 mole of CaCl2 is involved, the heat evolved is therefore:

Heat per mole = 384.09 J / 0.1225 mol

Heat = 3 135.43 J/mol

The enthalpy of solution is therefore 3153.43 J/mol or 3.15 kJ/mol.

5 0

2 years ago

If 25 g of NH3, and 96 g of H2S react according to the following reaction, what is the

jeyben [28]

25 g of NH₃ will produce 47.8 g of (NH₄)₂S

Explanation:

2 NH₃ + H₂S ----> (NH₄)₂S

Molecular weight of NH₃ = 17 g/mol

Molecular weight of (NH₄)₂S = 68 g/mol

According to the balanced reaction:

2 X 17 g of NH₃ produces 68 g of (NH₄)₂S

1 g of NH₃ will produce $\frac{68}{34}$ g of (NH₄)₂S

25g of NH₃ will produce $\frac{65}{34} X 25 g$ of (NH₄)₂S

= 47.8 g of (NH₄)₂S

Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S

4 0

2 years ago

A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (

djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

$Kb = \frac{[ HSO3-][0H-]}{SO3-2}$

$Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7$

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation $Kb=\frac{[HSO3-][OH-]}{[SO3-2]}$

We noe have:

$1.485*10^-^7=\frac{x*x}{(0.056-x)}$

$x = [OH-] = 9.11*10^-^5$

$pOH = -log(OH) = -log(9.11*10^-^5)$

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0

2 years ago

How many moles of PBr3 contain 3.68 x 10^25 bromine atoms?

scoray [572]

3.68 x 10²⁵ bromine atoms * 1mol/6.02*10²³ atoms=
= 61.13 mol of bromine atoms

1 mol PBr3 ----- 3 mol Br
x mol PBr3 -----61.13 mol Br

x= 1*61.13/3 = 20.4 mol PBr3.

20.4 mol PBr3 contain 3.68 x 10^25 bromine atoms.

7 0

2 years ago

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