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Irina-Kira [14]
3 years ago
6

Where is most of the high-level waste from nuclear reactors stored?

Chemistry
1 answer:
Nana76 [90]3 years ago
7 0
It's stored at the reactor site.
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blsea [12.9K]
Depending in the category of the Hurricane, you make experience different levels of wind power and destruction. Hurricanes only have 5 categories ranking from Category 1 to Category 5. The smallest category is category 1 making category 5 the largest. The bigger the category, the more wind or destruction you'll experience.
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3 years ago
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Where do you find an elements energy levels
ahrayia [7]
You can find an element's amount of energy level by determining their place on the periodic table. An element's amount of energy levels are represented by which period/ row they are in. For example, Calcium has 4 energy levels. I know this because it is in the fourth period on the table.

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3 years ago
Determine the percent water in MgSO4*7H20?
Semmy [17]

Answer:

\% H_2O=51.2\%

Explanation:

Hello!

In this case, since the percent water is computed by dividing the amount of water by the total mass of the hydrate; we infer we first need the molar mass of water and that of the hydrate as shown below:

MM_{MgSO_4* 7H_2O}=120.36 g/mol+7*18.02g/mol\\\\MM_{MgSO_4* 7H_2O}=246.5g/mol

Thus, the percent water is:

\% H_2O=\frac{7*MM_{H_2O}}{MM_{MgSO_4* 7H_2O}} *100\%\\\\

So we plug in to obtain:

\% H_2O=\frac{7*18.02}{246.5} *100\%\\\\\% H_2O=51.2\%

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5 0
3 years ago
Jumping on a cemented floor is more pain full than a sandy flour,why​
Anvisha [2.4K]

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Is soft

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5 0
3 years ago
60 grams of ice will require _____ calories to raise the temperature 1°C.
guajiro [1.7K]

Answer:

60 grams of ice will require 30.26 calories to raise the temperature 1°C.

Explanation:

The amount of heat (Q) to raise the temperature of 60.0 g of ice by 1°C can be calculated from:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of heat released or absorbed by the system.

m is the mass of the ice (m = 60.0 g).

c is the specific heat capacity of ice (c = 2.108 J/g.°C).

ΔT is the temperature difference (ΔT = 1.0 °C).

∴ Q = m.c.ΔT = (60.0 g)(2.108 J/g.°C)(1.0 °C) = 126.48 J.

<em>It is known that 1.0 cal = 4.18 J.</em>

<em>∴ Q = (126.48 J)(1.0 cal / 4.18 J) = 30.26 cal.</em>

7 0
3 years ago
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