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Llana [10]
2 years ago
7

6. The half life for uranium-235 is 7.0x108 years.

Chemistry
1 answer:
kaheart [24]2 years ago
4 0

Answer:

\large \boxed{4.0}

Explanation:

\text{No. of half-lives} = 2.8  \times 10^{9 } \text{ yr} \times \dfrac{\text{1 half-life}}{7.0 \times 10^{8} \text{ yr}} = \textbf{4.0 half-lives}\\\\\text{The sample went through $\large \boxed{\textbf{4.0 half-lives}}$}

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A turtle moves at a speed of 1.0km/h. How fast is the turtle moving in meter per second (m/s)?
ahrayia [7]

Answer:

0.28m/s

Explanation:

Speed is defined as the distance travelled per unit of time. The speed of the turtle is 1.0km/h. Thus, to find the speed in m/s, we need to convert km to m (1km is 1000m), and h to s (1h = 3600s).

<em>Converting units:</em>

1.0km/h * (1000m / 1km) * (1h / 3600s) = 0.28m/s.

The speed of the turtle in meter per second is 0.28m/s

7 0
2 years ago
What is the concentration of x2??? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5??10???6 and ka2=1.2??10???11
lutik1710 [3]
The first dissociation for H2X:
                        H2X +H2O ↔ HX + H3O
initial                0.15                     0      0
change             -X                     +X      +X
at equlibrium 0.15-X                  X        X
because Ka1 is small we can assume neglect x in H2X concentration
     Ka1      = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15) 
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
        HX + H2O↔ X^2 + H3O
    8.2x10^-4          Y         8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2       = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m


4 0
2 years ago
In a closed system, how will a decrease in pressure affect the following reaction: 2A(g) +2B(g) ⇌ 2C(g) + 2D(g)?
DochEvi [55]

As number of gaseous moles in reactant and prodict are same that is 4

So No change will occur

3 0
2 years ago
Read 2 more answers
Consider the decomposition of the compound C5H6O3 as follows below. C5H6O3(g) → C2H6(g) + 3 CO(g) When a 5.63-g sample of pure C
timofeeve [1]

Answer:

K = 6.5 × 10⁻⁶

Explanation:

C₅H₆O₃ ⇄ C₂H₆ + 3CO

Use PV=nRT to find the initial pressure of C₅H₆O₃

P (2.50) = (0.0493) (0.08206) (473)

P = 0.78atm

C₅H₆O₃ ⇄ C₂H₆ + 3CO

0.78atm      0           0

0.78 - x        x           3x

1.63atm = 0.78 - x + x + 3x

P(total) = 0.288atm

C₅H₆O₃ = 0.78 - 0.288

             = 0.489atm

C₂H₆ = 0.288atm

CO = 0.846atm

K_p = \frac{0.288 * 0.864^3}{0.489}

     = 0.379

K = \frac{K_p}{RT^3}

K = \frac{0.379}{(0.0821 * 473)^3}

   = 6.5 × 10⁻⁶

7 0
2 years ago
A company wants to start a nuclear power plant as a away to produce clean energy. They want to use the same reaction that's used
jok3333 [9.3K]

B.

The energy in nuclear fusion can't be generated in a controlled manner hence difficult to use in a nuclear power plant as a away to produce clean energy.

Explanation:

The reactions in thermonuclear bombs use very highly enriched radioactive elements that release a lot of energy at a go hence the enormous explosions they cause. A nuclear power plant would require this enormous energy to be released slowly  to generate electricity. It requires great technology to control the release of this nuclear energy because it requires the control of the rate of chain reactions occurring at a particular time.  If the chain reactions are too many at a time, the power plant could explode.

Learn More:

For more on nuclear energy check out;

brainly.com/question/1013815

#LearnWithBrainly

6 0
2 years ago
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