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2 years ago

50 Points (Brainliest too)! Please help!

Chemistry

1 answer:

solong [7]2 years ago

3 0

Answer:

Transmutation is the conversion of an atom of one element into an atom of a different element through nuclear changes.

An alpha particle is a particle composed of 2 protons and 3 neutrons. It results from transmutation because of the change in protons in a large nucleus.

A chemical equation is balanced accord to the number of atoms of each element before and after the change. This is also shows the Law of Conservation of Matter.

A nuclear equation is balanced according to mass number and charge (atomic number). These equations can’t be balanced like chemical equations because the identities of the atoms can change. On the left of equation, the top number is the atomic mass and the bottom number is the atomic number. Take note that if you add the atomic mass (top number), you’ll get the the atomic mass of of the atom pre-transmutation. Same applies with atomic number. This shows a balanced nuclear equation.

The atomic number is the number of protons in a nucleus. The mass number is the sum of the protons and neutrons in a nucleus. Electrons have very small masses, so they are not accounted in atomic mass.

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Salt breaks into a chlorine ion and sodium ion when dissolved in water. The chlorine ion has a negative charge. Which end of the

Monica [59]

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3 years ago

When 3.0 grams of H2 is reacted with excess C at constant pressure, the reaction forms CH4 and releases 53.3 kJ of heat. C(s) +

Murrr4er [49]

Answer:

THE ENTHALPY OF REACTION IN KJ/MOL OF CH4 IS 7.07 KJ/MOL.

Explanation:

Mass of H2 = 3 g

Molar mass of H2 = 2 g/mol

Heat released = 53.3 kJ

Equation of the reaction:

C(s) + 2H2(g) -------> CH4(g)

First:

Calculate the number of moles of H2 that was used:

Number of moles = mass / molar mass

Number of moles = 3g / 2g

Number of moles = 1.5 moles

So therefore, when 53.3 kJ of heat was released from the reaction, 1.5 moles of hydrogen was used.

From the equation of the reaction, one mole of carbon reacts with two moles of hydrogen to form one mole of methane.

For 3 g of hydrogen, 1.5 mole of hydrogen is involved.

It means:

1.5 moles of hydrogen reacts with 0.75 moles of carbon and produces 0.75 moles of methane. This is so because the reaction occurs in 1: 2: 1 in respect to carbon, hydrogen and methane respectively.

So we can say that the production of 0.75 mole of methane will evolve 53.3 kJ of heat.

0.75 mole of methane releases 53.3 kJ of heat.

1 mole of methane will release ( 53.3 kJ * 1 / 0.75 )

= 71.0666 kJ of heat

In conclusion, the enthalpy of the reaction in kJ/ mole of CH4 is 71.07 kJ/mol.

7 0

3 years ago

A 150.0 mL solution of 2.888 M strontium nitrate is mixed with 200.0 mL of a 3.076 M sodium fluoride solution. Calculate the mas

Lelechka [254]

Answer:

Mass SrF2 produced = 38.63 g SrF2 produced

[Na^+]: = 1.758 M

[NO3^-]: = 1.238 M

[Sr^2+] = 0.3589 M

[F^-] = 2.36*10^-5 M

Explanation:

Step 1: Data given

Volume of 2.888M strontium nitrate = 150.0 mL = 0.150 L

Volume of 3.076 M sodium fluoride = 200.0 mL = 0.200 L

Step 2 : The balanced equation

Sr(NO3)2(aq) + 2NaF(aq) → SrF2(s) + 2NaNO3(aq) → Sr2+ + 2F- + 2

Step 3: Calculate moles strontium nitrate

Moles Sr(NO3)2 = Molarity * volume

Moles Sr(NO3)2 = 2.888 M * 0.150 L

Moles Sr(NO3)2 = 0.4332 moles

Step 4: Calculate moles NaF

Moles NaF = 3.076 M * 0.200 L

Moles NaF = 0.6152 moles

It takes 2 moles F^- to precipitate 1 mole Sr^2+, so F^- is limiting.

Step 5: Calculate limiting reactant

For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3

NaF is the limiting reactant. It will completely be consumed (0.6152 moles).

Sr(NO3)2 is in excess. There will react 0.6152/2 = 0.3076 moles

Moles Sr^2+ precipitated by F^- = 0.3076

There will remain 0.4332 - 0.3076 = 0.1256 moles of Sr(NO3)2

Moles Sr^2+ no precipitated (left over) = 0.1256 moles

Step 6: Calculate moles SrF2

For 1 mol of Sr(NO3)2 we need 2 moles of NaF to produce 1 mol of SrF2 and 2 moles of NaNO3

For 0.6152 moles NaF we have 0.6152/2 = 0.3076 moles of SrF2

Mass SrF2 produced: 0.3076 mol * 125.6 g/mol = 38.63 g SrF2 produced

Step 7: Calculate concentration of [Na+] and [NO3-]

Since both Na^+ and NO3^- are spectator ions, and the final volume is 150 ml + 200 ml = 350 ml (0.350 L), the concentrations of Na^+ and NO3^- can be calculated as follows:

[Na^+]: (200 ml)(3.076 M) = (350 ml)(x M) and x = 1.758 M

[NO3^-]: (150 ml)(2.888 M)(2) = (350 ml)(x M) = 1.238 M

Step 8: Calculate [Sr^2+] and [F^-]

[Sr^2+] = 0.1256 moles/0.350 L = 0.3589 M

To find [F^-], one needs the Ksp for SrF2. There are several values listed in the literature. I am using a value of 2x10^-10.

SrF2(s) <==> Sr^2+(aq) + 2F^-(aq)

Ksp = [Sr^2+][F^-]²

2x10^-10 = (0.3589)(x)²

x² = 5.57*10^-10

x = [F^-] = 2.36*10^-5 M

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3 years ago

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topjm [15]

Answer:

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2 years ago

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