The chain reaction in nuclear reactors requires the absorption of neutrons in order to maintain control of the reaction.

The atomic weight of iodine is less than the atomic weight of tellurium. However, Mendeleev listed iodine after tellurium in his

ziro4ka [17]

Answer:

The reason for the suspicion was because the manner in which iodine reacted chemically as well as its other chemical properties, indicated that it belonged in the same group as chlorine and bromine, while the much heavier tellurium should be placed in the previous group

The suspicion was proved to be correct when the atomic number of tellurium was found to be 52 and that of iodine was found to be 53 by later scientists

Explanation:

5 0

4 years ago

Compared to other minerals, a mineral that is ranked higher on the moh’s scale will be:

skad [1K]

The choices here are:
<span> a. harder c. more dense b. shinier d. more reactive

The answer is "harder".
The Moh's Scale is also known as the </span><span>Mohs scale of mineral hardness. This scale is basin gon the ability of one certain material to visibly scratch another material. 1 Talc 2 Gypsum 3 Calcite 4 Fluorite 5 Apatite 6 Feldspar 7 Quartz 8 Topaz 9 Corundum 10 Diamond.</span>

3 0

3 years ago

A cell membrane acts as a boundary between the cell and its external environment. What is another role of the cell membrane?

Naddika [18.5K]

The answer is D, it controls what comes in & out the cell, well i hope so!

4 0

2 years ago

Read 2 more answers

The activation energy for the gas phase isomerization of dimethyl citraconate is 105 kJ.

Mamont248 [21]

Answer:

$k_2=9.06x10^{-4}/s$

Explanation:

Hello there!

In this case, since the activation energy, rate law and temperature, when variable, are related to each other as shown below:

$ln(\frac{k_2}{k_1} )=\frac{-Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )$

Thus, when solving for the rate constant at 682 K, we will obtain:

$ln(\frac{k_2}{2.77x10^{-4}/s} )=\frac{-105000J/mol}{8.3145\frac{J}{mol*K}}(\frac{1}{682K} -\frac{1}{641K} ) \\\\ln(\frac{k_2}{2.77x10^{-4}/s} )=1.184\\\\\frac{k_2}{2.77x10^{-4}/s}=exp(1.184)\\\\k_2=2.77x10^{-4}/s*3.269\\\\k_2=9.06x10^{-4}/s$

Best regards!

8 0

3 years ago

Purification of chromium can be achieved by electrorefining chromium from an impure chromium anode onto a pure chromium cathode

xenn [34]

Explanation:

It is given that mass is 11 kg. Convert mass into grams as follows.

$11 kg \times \frac{1000 g}{1 kg}$

= 11000 g (as 1 kg = 1000 g)

Now, calculate the number of moles as follows.

No. of moles = $\frac{\text{Mass of Cr}}{\text{Molar mass of Cr}}$

= $\frac{11000 g}{52 g/mol}$

= 211.54 mol

For $Cr^{3+}$ , 3 moles of electrons are required

Hence, $3 \times 211.54 mol$

= 634.62 mol

As 1 mol of electrons contain 96500 C of charge. Therefore, charge carried by 634.62 mol of electrons will be calculated as follows.

Q = $634.62 mol \times 96500 C/mol
$

= $61.24 \times 10^{6} C$ of charge

We know that relation between charge, current and time is as follows.

Q = $I \times t$

Current is given as 41.5 A and charge is calculated as $61.24 \times 10^{6} C$ . Therefore, calculate the time as follows.

Q = $I \times t$

$61.24 \times 10^{6} C = 41.5 A \times t$

t = $1.47 \times 10^{6} sec$

As there are 3600 seconds in one 1 hour. Therefore, converting $1.47 \times 10^{6} sec$ into hours as follows.

$\frac{1.47 \times 10^{6} sec}{3600 sec/hr}$

= 4.09 hr

Thus, we can conclude that it takes 4.09 hours to plate 11.0 kg of chromium onto the cathode if the current passed through the cell is held constant at 41.5 A.

6 0

3 years ago