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2 years ago

Two protons and two neutrons are released as a result of this reaction.

Chemistry

2 answers:

Ierofanga [76]2 years ago

6 0

Answer:

The answer is A

Explanation:

Murljashka [212]2 years ago

3 0

Answer: The particle released in the given reaction is one alpha particle

Explanation:

In a nuclear reaction, the total mass and total atomic number remains the same.

For the given fission reaction:

$^{222}_{86}\textrm{Rn}\rightarrow ^A_Z\textrm{X}+^{218}_{84}\textrm{Po}$

To calculate A:

Total mass on reactant side = total mass on product side

222 = A + 218

A = 4

To calculate Z:

Total atomic number on reactant side = total atomic number on product side

86 = Z + 84

Z = 2

The isotopic symbol of unknown element is $_{2}^{4}\textrm{He}$ . Another name of helium atom is alpha particle.

Hence, the particle released in the given reaction is one alpha particle

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The ph of a solution that contains 0.818 m acetic acid (ka = 1.76 x 10-5) and 0.172 m sodium acetate is __________.

crimeas [40]

PH is calculated using Handerson- Hasselbalch equation,

pH = pKa + log [conjugate base] / [acid]

Conjugate Base = Acetate (CH₃COO⁻)
Acid = Acetic acid (CH₃COOH)
So,
pH = pKa + log [acetate] / [acetic acid]

We are having conc. of acid and acetate but missing with pKa,
pKa is calculated as,

pKa = -log Ka
Putting value of Ka,
pKa = -log 1.76 × 10⁻⁵

pKa = 4.75
Now,
Putting all values in eq. 1,

pH = 4.75 + log [0.172] / [0.818]

pH = 4.072

4 0

2 years ago

1) Aluminum sulphate can be made by the following reaction: 2AlCl3(aq) + 3H2SO4(aq) Al2(SO4)3(aq) + 6 HCl(aq) It is quite solubl

kolezko [41]

Answer:

88.9%

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AlCl3(aq) + 3H2SO4(aq) —> Al2(SO4)3(aq) + 6HCl(aq)

Step 2:

Determination of the masses of AlCl3 and H2SO4 that reacted and the mass of Al2(SO4)3 produced from the balanced equation.

Molar mass of AlCl3 = 27 + (35.5x3) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar mass of Al2(SO4)3 = (27x2) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4 to produce 342g of Al2(SO4)3.

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4.

Therefore, 25g of AlCl3 will react with = (25 x 294)/267 = 27.53g of H2SO4.

From the calculations made above, we see that only 27.53g out 30g of H2SO4 given were needed to react completely with 25g of AlCl3.

Therefore, AlCl3 is the limiting reactant and H2SO4 is the excess.

Step 4:

Determination of the theoretical yield of Al2(SO4)3.

In this case we shall be using the limiting reactant because it will produce the maximum yield of Al2(SO4)3 since all of it is used up in the reaction.

The limiting reactant is AlCl3 and the theoretical yield of Al2(SO4)3 can be obtained as follow:

From the balanced equation above,

267g of AlCl3 reacted to produce 342g of Al2(SO4)3.

Therefore, 25g of AlCl3 will react to produce = (25 x 342) /267 = 32.02g of Al2(SO4)3.

Therefore, the theoretical yield of Al2(SO4)3 is 32.02g

Step 5:

Determination of the percentage yield of Al2(SO4)3.

This can be obtained as follow:

Actual yield of Al2(SO4)3 = 28.46g

Theoretical yield of Al2(SO4)3 = 32.02g

Percentage yield of Al2(SO4)3 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 28.46/32.02 x 100

Percentage yield = 88.9%

Therefore, the percentage yield of Al2(SO4)3 is 88.9%

3 0

2 years ago

analysis of a compound indicates that it contains 1.04 grams K 0.70 g Cr and 0.86 g O. Find its empirical formula

MrMuchimi

1.04gK*1molK/39.01g K= 0.0267 mol K
0.70gCr*1mol/52.0g Cr = 0.0135 mol Cr
0.86 gO* 1 mol/16.0 g O = 0.0538 mol O
0.0267 mol K/0.0135 = 2 mol K
0.0135 mol Cr /0.0135= 1 mol Cr
0.0538 mol O/0.035= 4 mol Cr
K2CrO4

6 0

2 years ago

Read 2 more answers

What is the emperical formula for a compound containing 68.3% lead, 10.6% sulfur, and the remainder oxygen? a. Pb2SO4 b. PbSO3 c

brilliants [131]

Answer:

Molecular formula → PbSO₄ → Lead sulfate

Option c.

Explanation:

The % percent composition indicates that in 100 g of compound we have:

68.3 g of Pb, 10.6 g of S and (100 - 68.3 - 10.6) = 21.1 g of O

We divide each element by the molar mass:

68.3 g Pb / 207.2 g/mol = 0.329 moles Pb

10.6 g S / 32.06 g/mol = 0.331 moles S

21.1 g O / 16 g/mol = 1.32 moles O

We divide each mol by the lowest value to determine, the molecular formula

0.329 / 0.329 = 1 Pb

0.331 / 0.329 = 1 S

1.32 / 0.329 = 4 O

Molecular formula → PbSO₄ → Lead sulfate

8 0

2 years ago

Read 2 more answers

Which mass of gas would occupy a volume of 3dm3 at 25°C and 1 atmosphere pressure?

balu736 [363]

Answer:

C 8.09 SO2 gas

Explanation:

As we have the volume (3dm³ = 3L), temperature (25°C + 273 = 298K), and pressure (1atm), we can solve to moles of gas using:

PV = nRT

PV / RT = n

1atm*3L / 0.082atmL/molK*298K =¨

0.123 moles of gas you have.

Now, to convert these moles to mass we use molar mass (32g/mol for O2, 28g/mol for N2, 64g/mol for SO2, and 44g/mol for CO2).

Mass of 0.123 moles of these gases is:

O2 = 0.123 moles * 32g/mol = 3.94g of O2. A is wrong

N2 = 0.123 moles * 28g/mol = 3.4g of N2. B is wrong

SO2 = 0.123 moles * 64.1g/mol = 7.9g of SO2≈ 8.09g of SO2, C is possible

CO2 = 0.123 moles * 44g/mol = 5.4g of CO2. D is wrong

Right answer is:

8 0

2 years ago