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zubka84 [21]
3 years ago
8

While working near a radioactive mine, workers usually wear heavy plastic suits. These suits protect them from all but

Chemistry
2 answers:
Vaselesa [24]3 years ago
4 0
The appropriate response is gamma radiation. Alpha particles can be halted via air. UV radiation can be halted by a typical layer of clothing.Beta particles can be ceased by the thick plastic suit. Just gamma radiation can enter the substantial suit. It must be halted by thick dividers of lead or cement.
emmasim [6.3K]3 years ago
4 0

gamma radiation would be the best answer. Alpha particles can be stopped by air. UV radiation can be halted by a layer of clothing or insulator. Beta particles can be ceased by the plastic suit. Just gamma radiation can enter the substantial suit. It must be halted by thick dividers of lead or cement.

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calculate the volume that will be occupied by 350 ml of oxygen measured at 720 mm hg, when the pressure changed to 630 mm hg
likoan [24]

Answer:

406.45mL

Explanation:

The following data were obtained from the question:

V1 = 350mL

P1 = 720mmHg

P2 = 630mmHg

V2 =?

The new volume can be obtain as follows:

P1V1 = P2V2

720 x 350 = 620 x v2

Divide both side by 620

V2 = (720 x 350) /620

V2 = 406.45mL

The new volume of the gas is 406.45mL

8 0
3 years ago
Calculate Ho298 for the process
Inga [223]

Explanation:

As per the Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

Hence, according to this law the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

Sb + \frac{3}{2}Cl_2 \rightarrow SbCl_{3}    \Delta H^0_1 =  -314 kJ  ..........(1)

SbCl_{3} + Cl_2 \rightarrow SbCl_{5}    \Delta H^0_2 = -80kJ   ..............(2)

The final reaction is as follows:  

Sb + \frac{5}{2}Cl_{2} \rightarrow SbCl_{5}  \Delta H^0_3 = ?  .............(3)

Therefore, adding (1) and (2) we get the final equation (3) and value of \Delta H^{0}_{3} at 298 K will be as follows.

             \Delta H^{0}_{3} = \Delta H^{0}_{1} + \Delta H^{0}_{2}    

                       = -314 kJ + (-80) kJ

                       = -394 kJ

Thus, we can conclude that H^{o} at 298 K for the given process is -394 kJ.

4 0
3 years ago
Determine the number of formula units in 48.0 grams of magnesium chloride (MgCl2)
vekshin1

Answer:

3.03 x 10²³ formula units

Explanation:

First we <u>convert 48.0 grams of magnesium chloride into moles</u>, using its <em>molar mass</em>:

  • 48.0 g ÷ 95.21 g/mol = 0.504 mol MgCl₂

Then we <u>convert 0.504 moles into formula units</u>, using <em>Avogadro's number</em>:

  • 0.504 mol * 6.023x10²³ formula units/mol = 3.03x10²³ formula units
3 0
3 years ago
Last 10 Year Noble Prize Winners in Chemistry from India​
telo118 [61]

Venkatraman Ramakrishan

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8 0
3 years ago
Calculate the mass of 0.100 mol of calcium
miv72 [106K]

Answer:

4.01 g

Explanation:

7 0
3 years ago
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