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4 years ago

Select the major problems associated with the production of nuclear energy

Chemistry

1 answer:

zhuklara [117]4 years ago

3 0

Nuclear energy = nuclear bobm = nuclear war = bad ;(

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The value for ψ in root tissue was found to be -1.6 bars. If you place the root tissue in a 0.1 m solution of sucrose at 20°c in

Allisa [31]

Answer:

$\Psi=\Psi_s+\Psi_p\\$

$I=1;C=0.1;R=0.0831\\$

$T=20+273=293~C\\$

as beaker is open so

$\Psi_p=0\\$

$\Psi=\Psi_s\\\Psi=-2.43483$

As ψof solution is lower than the root tissue so water will move out of the root to the solution

Explanation:

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4 years ago

Devices that allow you to do work in an easier way.

IrinaVladis [17]

Answer:

The most notable of these are known as the "six simple machines": the wheel and axle, the lever, the inclined plane, the pulley, the screw, and the wedge, although the latter three are actually just extensions or combinations of the first three.

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3 years ago

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How many electrons are transferred when magnesium bonds ionically with oxygen

andrew11 [14]

Answer:

2 electrons from magnesium (Mg) is transferred to oxygen (O2) to form Mg2+ and O2-

which combines by strong electrostatic force to form MgO

the equation is

2Mg + O2 -> 2MgO

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3 years ago

What is the correct formula for Strontium Selenide?

ikadub [295]

Answer:

(SrSe) | SeSr - PubChem.

Explanation:

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3 years ago

A chunk of tin weighing 18.5 grams and originally at 97.38 °C is dropped into an insulated cup containing 75.7 grams of water at

weqwewe [10]

Answer:

22.44°C will be the final temperature of the water.

Explanation:

Heat lost by tin will be equal to heat gained by the water

$-Q_1=Q_2$

Mass of tin = $m_1=18.5 g$

Specific heat capacity of tin = $c_1=0.21 J/g^oC$

Initial temperature of the tin = $T_1=97.38^oC$

Final temperature = $T_2$ =T

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2=75.7 g$

Specific heat capacity of water= $c_2=4.184 J/g^oC$

Initial temperature of the water = $T_3=21.52^oC$

Final temperature of water = $T_2$ =T

$Q_2=m_2c_2\times (T-T_3)$

$-Q_1=Q_2$

$-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)$

On substituting all values:

$-(18.5 g\times 0.21 J/g^oC\times (T-97.38^oC))=75.7 g\times 4.184 J/g^oC\times (T-21.52 ^oC)$

we get, T = 22.44°C

22.44°C will be the final temperature of the water.

5 0

4 years ago