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4 years ago

8

Danial has a simple of pure copper.its mass 89.6 grams (g),and its volume is 10 cubic centermeters (cm3) whats the destiny of th

e sample?

1 answer:

drek231 [11]4 years ago

6 0

Answer:

8.96g\ cm3

Explanation:

D = ( 89.6g \ 10cm3)

( 89.6\ 10) ( g\ cm3) = 8.96g\cm3

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How much potassium iodide to take for radiation protection.

Ket [755]

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Explanation:

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3 years ago

What are some safety precautions that must be taken in order for nuclear reactions to occur? Give at least 5 precautions

rodikova [14]

Answer:

1. Control of Radioactivity

This requires being able to control the neutron flux. Recall that in a nuclear reactor when a neutron is captured by a fuel nucleus (generally uranium) the nucleus splits releasing radioactive particles (or undergoes fission). Hence if we decrease the neutron flux we decrease the radioactivity. The most common way to reduce the neutron flux is include neutron-absorbing control rods. These control rods can be partially inserted into the reactor core to reduce the reactions. The control rods are very important because the reaction could run out of control if fission events are extremely frequent. In modern nuclear power plants, the insertion of all the control rods into the reactor core occurs in a few seconds, thus halting the nuclear reaction as rapidly as possible. In addition, most reactors are designed so that beyond optimal level, as the temperature increases the efficiency of reactions decreases, hence fewer neutrons are able to cause fission and the reactor slows down automatically.

2. Maintenance of Core Cooling

In any nuclear reactor some sort of cooling is necessary. Generally nuclear reactors use water as a coolant. However some reactors which cannot use water use sodium or sodium salts.

3. Maintenance of barriers that prevent the release of radiation

There is a series of physical barriers between the radioactive core and the environment. For instance at the Darling Nuclear Generation Station in Canada the reactors are enclosed in heavily reinforced concrete which is 1.8m thick. Workers are shielded from radiation via interior concrete walls. A vacuum building is connected to the reactor buildings by a pressure relief duct. The vacuum building is a 71m high concrete structure and is kept at negative atmospheric pressure. This means that if any radiation were to leak from the reactor it would be sucked into the vacuum building and therefore prevented from being released into the environment.

The design of the reactor also includes multiple back-up components, independent systems (two or more systems performing the same function in parallel), monitoring of instrumentation and the prevention of a failure of one type of equipment affecting any other.

Further, regulation requires that a core-meltdown incident must be confined only to the plant itself without the need to evacuate nearby residence.

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The handling of equipment via remote in the core of the reactor

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3 years ago

Which of these changes could be classified as producing a chemical changes

podryga [215]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

If 72.5 grams of calcium metal (Ca) react with 65.0 grams of oxygen gas (O2) in a synthesis reaction, how many grams of the exce

Natali5045456 [20]

2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)

65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.

Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)

=0.0282669621 g of O2 left over

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4 years ago

The reaction of 9.50 g of carbon with Excess O2 Yield 12.8 of CO2. What is the percent yield of this reaction?

Anna [14]

Answer:

Percentage yield = 36.75%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

C + O₂ —> CO₂

Next, we shall determine the mass of C that reacted and the mass of CO₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of C = 12 g/mol

Mass of C from the balanced equation = 1 × 12 = 13 g

Molar mass of CO₂ = 12 + (2×16)

= 12 + 32

= 44 g/mol

Mass of CO₂ from the balanced equation = 1 × 44 = 44 g

SUMMARY:

From the balanced equation above,

12 g of C reacted to produce 44 g of CO₂.

Next, we shall determine the theoretical yield of CO₂. This can be obtained as follow:

From the balanced equation above,

12 g of C reacted to produce 44 g of CO₂.

Therefore, 9.50 g of C will react to produce = (9.50 × 44) / 12 = 34.83 g of CO₂.

Thus, the theoretical yield of CO₂ is 34.83 g.

Finally, we shall determine the percentage yield of the reaction. This can be obtained as follow:

Actual yield of CO₂ = 12.8

Theoretical yield of CO₂ = 34.83 g

Percentage yield =?

Percentage yield = Actual yield / Theoretical yield × 100

Percentage yield = 12.8 / 34.83 × 100

Percentage yield = 1280 / 34.83

Percentage yield = 36.75%

5 0

3 years ago