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natka813 [3]
3 years ago
10

When radium-226 (atomic number-88) decays by emitting an alpha particle it becomes _____.

Chemistry
1 answer:
topjm [15]3 years ago
3 0

Answer:

d. Radon-222

Explanation:

²²⁶₈₈Ra → ²²²₈₆Rn  + ⁴₂He

Alpha particle is a helium nucleus with mass number 4 and atomic number 2. According to the law of conversation of mass, the sum of the mass number and atomic number must be equal on both side of the reaction.  

Since the mass number of Ra is 226 and that of He is 4. The mass number of the unknown element must be 226 - 4 = 222.  

Since the atomic number of Ra is 88 and that of He is 2. The atomic number of the unknown element must be 88 - 2 = 86.  

Now looking in the periodic table Radon is the only element with atomic number 86.  

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How many hydrogen atoms are in 3 molecules of ethyl alcohol, C2H3OH
Andrew [12]

Answer: 3 hydrogen atoms

Explanation: H=hydrogen

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For which of the following processes would you expect there to be an increase in entropy? Ag+(aq) + Cl-(aq) AgCl(s) H2O(g) H2O(l
Lyrx [107]
CaBr₂(s) → Ca⁺²(aq) + 2Br⁻ (aq)     ΔS>0
6 0
3 years ago
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1.65g of zinc is used to make 8g of zinc iodide. How much iodine is required for this reaction?
creativ13 [48]

Answer:

6.45 g of iodine, I₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

Zn + I₂ —> ZnI₂

Next, we shall determine the mass of Zn and I₂ that reacted from the balanced equation. This can be obtained as follow:

Molar mass of Zn = 65 g/mol

Mass of Zn from the balanced equation = 1 × 65 = 65 g

Molar mass of I₂ = 127 × 2 = 254 g/mol

Mass of I₂ from the balanced equation = 1 × 254 = 254 g

SUMMARY:

From the balanced equation above,

65 g of Zn reacted with 254g of I₂.

Finally, we shall determine the mass of f I₂ needed to react with 1.65 g of Zn. This can be obtained as follow:

From the balanced equation above,

65 g of Zn reacted with 254g of I₂.

Therefore, 1.65 g of Zn will react with = (1.65 × 254)/65 = 6.45 g of I₂.

Thus, 6.45 g of iodine, I₂ is needed for the reaction.

4 0
3 years ago
The molar mass is determined by measuring the freezing point depression of an aqueous solution. A freezing point of -5.20°C is r
Dima020 [189]

Answer:

The empirical formula is C2H4O3

The molecular formula is C4H8O6

The molar mass is 152 g/mol

Explanation:

The complete question is: An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of -5.20°C is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Step 1: Data given

Mass % of Carbon = 31.57 %

Mass % of H = 5.30 %

Freezing point = -5.20 °C

10.56 grams of the compound dissolved in 25.0 grams of water

Kf water = 1.86 °C kg/mol

Step 2: Calculate moles of Carbon

Suppose 31.57% = 31.57 grams

moles C = mass C / Molar mass C

moles C = 31.57 grams / 12.0 g/mol = 2.63 moles

Step 3: Calculate moles of Hydrogen:

Moles H = 5.30 grams / 1.01 g/mol

moles H = 5.25 moles

Step 4: Calculate moles of Oxygen

Moles O = ( 100 - 31.57 - 5.30) / 16 g/mol

Moles O = 3.95 moles

Step 5: We divide by the smallest number of moles

C: 2.63 / 2.63 = 1 → 2

H: 5.25/2.63 = 2 → 4

O: 3.95/ 2.63 = 1.5 → 3

The empirical formula is C2H4O3

The molar mass of the empirical formula = 76 g/mol

Step 6: Calculate moles solute

Freezing point depression = 5.20 °C = m * 1.86

m = 5.20 / 1.86

m = 2.80 molal = 2.80 moles / kg

2.80 molal * 0.025 kg = 0.07 moles

Step 7: Calculate molar mass

Molar mass = mass / moles

Molar mass = 10.56 grams / 0.07 moles

Molar mass = 151 g/mol

Step 8: Calculate molecular formula

151 / 76 ≈  2

We have to multiply the empirical formula by 2

2*(C2H4O3) = C4H8O6

The molecular formula is C4H8O6

The molar mass is 152 g/mol

6 0
3 years ago
How much heat is absorbed/released when 20.00 g of NH3(g) reacts in the presence of excess O2 (g) to produce NO (g) and H2O (l)
FrozenT [24]

Answer:

a. 342.9 kJ of heat are absorbed.

Explanation:

Calculation of the moles of NH_3 as:-

Mass = 20.00 g

Molar mass of NH_3 = 17.031 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{20.00\ g}{17.031\ g/mol}

Moles= 1.1743\ mol

Given that:- \Delta H=+1168\ kJ

It means that 1 mole of NH_3 undergoes reaction and absorbs 1168\ kJ of heat

So,

1168 mole of NH_3 undergoes reaction and absorbs 1168\times 1168\ kJ of heat

<u>Amount of heat absorbed = + 342.9 KJ</u>

7 0
3 years ago
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