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dolphi86 [110]
3 years ago
7

When a certain nuclide undergoes alpha emission, astatine-217 is produced. What is the identity of the nuclide that underwent de

cay?
a. actinium-219
b. francium-217
c. francium-221
d. astatine-221
e. actinium-221
Please explain the answer
Chemistry
1 answer:
Olin [163]3 years ago
8 0

Answer:

C: francium-221

Explanation:

First of all to get a broader perspective, every isotope of francium usually undergoes decay to form astatine, radium, or radon.

Now, Francium-223 and francium-221 are it's only isotopes that occur in nature.

However, francium-221 is the one that undergoes alpha decay to produce astatine-217.

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A container of gas has a volume of 3.5 L and a pressure of 0.8 atm. Assuming the temperature remains constant, what volume of ga
strojnjashka [21]

Answer:

The correct answer is option A.

Explanation:

Initial volume of the gas =V_1=3.5 L

Final volume of the gas = V_2=?

Initial pressure of the gas =P_1=0.8 atm

Final volume of the gas = P_2=0.5 atm

Using Boyle's law:

P_1V_1=P_2V_2

0.8 atm \times 3.5 L=0.5 atm\times V_2

V_2=5.6 L

Hence,the correct answer is option A.

7 0
3 years ago
Read 2 more answers
Physical science!!! please help
IgorLugansk [536]

Answer:

The correct answer is : No, because there are 4 hydrogen atoms on the reactants side and 2 on the products side.

Explanation:

NH_4NO_3\rightarrow N_2O+H_2O

The given reaction equation is not balanced because:

  • Number of hydrogen atoms  on both sides are not equal that is 4 on reactants side and 2 on products side.
  • Number of oxygen atoms on both sides are not equal that is 3 on reactants side and 2 on products side.

In a balanced chemical equation number of atoms of each elements are equal on both sides.

So, the balanced chemical equation will be:

NH_4NO_3\rightarrow N_2O+2H_2O

7 0
3 years ago
Calculate the volume of a sample of mercury with a density of 14.6 g/mL and a mass of 1.00 g. The answer is assumed to be in mL.
Darya [45]

Answer:

0.0685 mL

Explanation:

To find the volume of the sample, divide the mass by the density.

(1.00 g)/(14.6 g/mL) = 0.0685 mL

3 0
3 years ago
Analysis of skunk spray yields a molecule with 44.77% c, 7.46% h and 47.76% s by mass. what is the empirical formula for this mo
stiv31 [10]
Suppose we have 100 gr of the substance. Then by weight, it would contain 44.77 gr of C, 7.46 gr of H and 47.76 gr of S. We need to look up the atomic weights of these atoms; M_H=1, M_C=12, M_S=32. The following formula holds (where n are the moles of the substance, M its molecular mass and m its mass): n=m/M. Substituting the known quantities for each element, we get that the substance has 3.73 moles of C, 7.46 moles of H and 1.49 moles of S. In the empirical formula for the molecule, all atoms appear an integer amout of times. Hence, for every mole of Sulfur, we have 2.5 moles of C and 5 moles of H (by taking the moles ratios). Thus, for every 2 moles of sulfur, we have 5 moles of C and 10 moles of H. Now that all the coefficients are integer, we have arrived at an empirical formula for the skunk spray agent: C_5H_{10}S_2
4 0
4 years ago
Suppose a 250. mL flask is filled with 0.30 mol of N_2 and 0.70 mol of NO. The following reaction becomes possible:N_2(g) +O2 →
Inessa [10]

Answer:

0.4 M

Explanation:

Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.

Because there's no O₂ in the beginning, the NO will decompose:

N₂(g) + O₂(g) ⇄ 2NO(g)

0.30 0 0.70 Initial

+x +x -2x Reacts (the stoichiometry is 1:1:2)

0.30+x x 0.70-2x Equilibrium

The equilibrium concentrations are the number of moles divided by the volume (0.250 L):

[N₂] = (0.30 + x)/0.250

[O₂] = x/0.25

[NO] = (0.70 - 2x)/0.250

K = [NO]²/([N₂]*[O₂])

K = \frac{(\frac{0.70 -2x}{0.250})^2 }{\frac{0.30+x}{0.250}*\frac{x}{0.250} }

7.70 = (0.70-2x)²/[(0.30+x)*x]

7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)

4x² - 2.80x + 0.49 = 2.31x + 7.70x²

3.7x² + 5.11x - 0.49 = 0

Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70

x = 0.09 mol

Thus,

[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M

3 0
3 years ago
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