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3 years ago

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Given the balanced equation below, how many moles of chromium are

1 answer:

fenix001 [56]3 years ago

6 0

Bbbbbbbbbbbbbbbbbbbbbbbb

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What is the difference between natural and artificial transmutation? Give an<br> example of each.

cricket20 [7]

Explanation:

Natural transmutation is the spontaneous disintegration of a heavy nuclide into lighter ones or fusion of lighter nuclides into heavier ones. These processes occurs naturally without anything inducing them. Nuclear fission of light elements in the core of stars is an example of natural transmutation.

Typically, all nuclei with atomic number greater than 83 are naturally radioactive.

²³⁸₉₂U → ²³⁴₉₂Th + ⁴₂He

In artificial transmutation, nuclear reactions are initiated through the collision of a nuclide with a high speed particle. Here, unstable nuclei are produced artificially in nuclear reactions. Such unstable nuclei also produce radiations.

¹₀n + ²³₁₁Na → ²⁴₁₁Na + γ

Learn more:

Transmutation brainly.com/question/3433940

#learnwithBrainly

6 0

3 years ago

In order for convection to transfer heat, particles need to

inna [77]

It’s b make contact with the heat source

7 0

2 years ago

Read 2 more answers

Cl2 + NaBr<br>which two elements will trade places?

anygoal [31]

The hallogens chloride with br

5 0

3 years ago

A sample of gas contains 6.25 × 10-3 mol in a 500.0 mL flask at 265°C. What is the pressure of the gas in kilopascals? Which var

RideAnS [48]

55.9 kPa; Variables given = volume (V), moles (n), temperature (T)

We must calculate <em>p</em> from <em>V, n</em>, and <em>T</em>, so we use <em>the Ideal Gas Law</em>:

<em>pV = nRT</em>

Solve for <em>p</em>: <em>p = nRT/V</em>

R = 8.314 kPa.L.K^(-1).mol^(-1)

<em>T</em> = (265 + 273.15) K = 538.15 K

<em>V</em> = 500.0 mL = 0.5000 L

∴ <em>p</em> = [6.25 x 10^(-3) mol x 8.314 kPa·L·K^(-1)·mol^(-1) x 538.15 K]/(0.5000 L) = 55.9 kPa

4 0

3 years ago

Read 2 more answers

In a first-order decomposition reaction, 50.0% of a compound decomposes in 10.5 min.

mihalych1998 [28]

In this reaction 50% of the compound decompose in 10.5 min thus, it is half life of the reaction and denoted by symbol $t_{1/2}$ .

(a) For first order reaction, rate constant and half life time are related to each other as follows:

$k=\frac{0.6932}{t_{1/2}}=\frac{0.6932}{10.5 min}=0.066 min^{-1}$

Thus, rate constant of the reaction is $0.066 min^{-1}$ .

(b) Rate equation for first order reaction is as follows:

$k=\frac{2.303}{t_{1/2}}log\frac{[A_{0}]}{[A_{t}]}$

now, 75% of the compound is decomposed, if initial concentration $[A_{0} ]$ is 100 then concentration at time t $[A_{t} ]$ will be 100-75=25.

Putting the values,

$0.066 min^{-1}=\frac{2.303}{t}log\frac{100}{25}=\frac{2.303}{t}(0.6020)$

On rearranging,

$t=\frac{2.303\times 0.6020}{0.066 min^{-1}}=21 min$

Thus, time required for 75% decomposition is 21 min.

8 0

3 years ago