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3 years ago

What percentage of a radioactive species would be found as daughter material after six half–lives?

1 answer:

3 0

100%.....50%.....25%......12.5%......6.25%......3.125%......1.5625%
...........1............2...........3..............4.................5................6..................

After six half-lives would be found 1.5625% of readioactive species.

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A copper ore containing 4.3 wt% water, 10.3 wt% CuSO, and 85.3 wt% rock is fed to a staged counter current extractor at a rate o

Serga [27]

Answer:

6 stages

Explanation:

Given:

Copper ore contains:

4.3 wt% water

10.3 wt% CuSO₄

85.3 wt% rock

Feed rate = 11.7 tons per hour

solution produced is to be 10 wt% copper sulfate

90 wt% water

Now,

The CuSO₄ supplied through the feed = 10.3 wt% CuSO₄ × Feed rate

= 0.103 × 11.7

= 1.2051 tons per hour

also,

At each stage 2 tons of solution is collected

therefore,

The CuSO₄ collected in each stage

= weight of solution × 10 wt% copper sulfate

= 2 tons per hour × 0.1

= 0.2 tons per hour

also,

98% of the CuSO₄ is to be recovered

thus,

the amount of CuSO₄ to be recovered

= 0.98 × CuSO₄ supplied through the feed

= 0.98 × 1.2051 tons per hour

= 1.180998

Therefore,

The number of stages required = $\frac{\textup{amount of CuSO₄ to be recovered }}{\textup{CuSO₄ collected in each stage }}$

= $\frac{\textup{1.180998}}{\textup{0.2}}$

= 5.90499 ≈ 6 stages

8 0

3 years ago

The first excited electronic energy level of the helium atom is 3.13 x 10-18 J above the ground level. Estimate the temperature

salantis [7]

Answer:

75603.86473 K

Explanation:

Given that:

The 1st excited electronic energy level of He atom = 3.13 × 10⁻¹⁸ J

The objective of this question is to estimate the temperature at which the ratio of the population will be 5.0 between the first excited state to the ground state.

The formula for estimating the ratio of population in 1st excited state to the ground state can be computed as:

$\dfrac{N_2}{N_1} = e ^{^{-\dfrac{(E_2-E_1)}{KT}}} = e ^{^{-\dfrac{(\Delta E)}{KT}}}$

From the above equation:

Δ E = energy difference = 3.13 × 10⁻¹⁸ J

k = Boltzmann constant = 1.38 × 10⁻²³ J/K

$\dfrac{N_2}{N_1} = 0.5$

Thus:

$0.05 =e^{^{ -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}$

$In (0.05) = { -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}$

$-3.00 = { -\dfrac{3.13 \times 10^{-18} \ J}{1.38\times 10^{-23 \ J/K}\times T}}}$

$-3.00 = -226811.5942 \times \dfrac{1}{T}$

$T = \dfrac{-226811.5942}{-3.00 }$

T = 75603.86473 K

4 0

3 years ago

A compound that changes color when it is in contact with an acid or base is called

malfutka [58]

It's called a pH indicator.

4 0

3 years ago

Predict the sign for AS for each of the following systems: Water freezing Water evaporating Crystalline urea dissolving Assembly

djyliett [7]

Answer: 1. Water freezing : $\Delta S$ is -ve.

2. Water evaporating : $\Delta S$ is +ve.

3. Crystalline urea dissolving : $\Delta S$ is +ve.

4. Assembly of the plasma membrane from individual lipids: $\Delta S$ is -ve.

5. Assembly of a protein from individual amino acids: $\Delta S$ is -ve.

Explanation:

Entropy is defined as the measurement of degree of randomness in a system.

It is represented by symbol S and we can only measure a change in entropy which is given by $\Delta S$ .

If there is decrease in randomness , the sign for $\Delta S$ is -ve and If there is increase in randomness , the sign for $\Delta S$ is +ve.

1. Water freezing: Entropy decreases as we move from liquid state to to solid state and thus $\Delta S$ is -ve.

2. Water evaporating : Entropy increases as we move from liquid state to gaseous state and thus $\Delta S$ is +ve.

3. Crystalline urea dissolving : The molecules convert from solid and ordered state to aqueous phase and random state. Thus the entropy increases and thus $\Delta S$ is +ve.

4. Assembly of the plasma membrane from individual lipids: random lipids are associating to form a single large polymer and thus entropy decreases and thus $\Delta S$ is -ve.

5. Assembly of a protein from individual amino acids: random amino acids are associating to form a single large polymer and thus entropy decreases and thus $\Delta S$ is -ve.

5 0

3 years ago

If boron atoms have 5 electrons, how many electrons are in the outer electron shell of a boron atom?

ANEK [815]

Answer:

Explanation:

The first election shell can only hold 2 electrons, but the next one can hold up to 8

3 0

2 years ago