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2 years ago

7

What are radioactive isotopes?

1 answer:

yan [13]2 years ago

3 0

Answer:

A radioactive isotope, also known as a radioisotope, radionuclide, or radioactive nuclide, is any of several species of the same chemical element with different masses whose nuclei are unstable and dissipate excess energy by spontaneously emitting radiation in the form of alpha, beta, and gamma rays.

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When equal moles of an acid and a base are mixed, after reaction the two are compounds are said to be at the _______________. Se

Elodia [21]

Answer:

when equal moles of an acid and base are mixed,after reaction the two are compounds are said to be at the Equivalent point.

3 0

2 years ago

It is not f so it could be anyother answer

uysha [10]

H is the answer :)
Hope that helps

4 0

3 years ago

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Liquid helium boils at –268.93 °C. What is the boiling point of helium on the Kelvin temperature scale?

BARSIC [14]

In order to change celcius to kelvin always add 73 to it leaving you with -195.93

4 0

2 years ago

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what volume of a 0.155 M calcium hydroxide solution is required to neutralize 28.8 mL of a 0.106 M nitric acid?

Hitman42 [59]

Answer:

9.85mL

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below:

Ca(OH)2 + 2HNO3 —> Ca(NO3)2 + 2H2O

From the balanced equation above,

nA (mole of the acid) = 2

nB (mole of the base) = 1

Data obtained from the question include:

Vb (volume of the base) =?

Mb (Molarity of base) = 0.155 M

Va (volume of the acid) = 28.8 mL

Ma (Molarity of acid) = 0.106 M

Using MaVa/MbVb = nA/nB, the volume of calcium hydroxide (i.e the base) can be obtain as follow:

MaVa/MbVb = nA/nB

0.106 x 28.8 / 0.155 x Vb = 2/1

Cross multiply to express in linear form as shown below:

2 x 0.155 x Vb = 0.106 x 28.8

Divide both side by 2 x 0.155

Vb = (0.106 x 28.8) / (2 x 0.155)

Vb = 9.85mL

Therefore, the Volume of calcium hydroxide is 9.85mL

7 0

2 years ago

A 2.04 g lead weight, initially at 10.8 oC, is submerged in 7.62 g of water at 52.3 oC in an insulated container. clear = 0.128

alisha [4.7K]

Answer: The final temperature of both the weight and the water at thermal equilibrium is $50.26^{o}C$ .

Explanation:

The given data is as follows.

mass = 7.62 g, $T_{2} = 10.8^{o}C$

Let us assume that T be the final temperature. Therefore, heat lost by water is calculated as follows.

q = $mC \times \Delta T$

= $7.62 g \times 4.184 J/^{o}C \times (52.3 - T)$

Now, heat gained by lead will be calculated as follows.

q = $mC \times \text{Temperature change of lead}$

= $2.04 \times 0.128 \times (T - 11.0)$

According to the given situation,

Heat lost = Heat gained

$7.62 g \times 4.184 J/^{o}C \times (52.3 - T)$ = $2.04 \times 0.128 \times (T - 11.0)$

T = $50.26^{o}C$

Thus, we can conclude that the final temperature of both the weight and the water at thermal equilibrium is $50.26^{o}C$ .

7 0

3 years ago