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2 years ago

Find the number of moles of the following substances 210 grams of NaHCO3

Chemistry

1 answer:

Triss [41]2 years ago

8 0

The number of moles of 210 grams of NaHCO₃ is 2.5 moles.

<h3>How to find the Number of moles ? </h3>

To calculate the number of moles use the formula

Number of moles = $\frac{\text{Given Mass}}{\text{Molar Mass}}$

Mass of NaHCO₃ = 210 g

Now we have to find the Molar mass of NaHCO₃

= Atomic mass of Na + Atomic mass of H + Atomic mass of C + 3 (Atomic mass of O)

= 23 + 1 + 12 + 3 (16)

= 36 + 48

= 84 g/mol

Now put the value in above formula we get

Number of moles = $\frac{\text{Given Mass}}{\text{Molar Mass}}$

= $\frac{210\ g}{84\ g/mol}$

= 2.5 moles

Thus from the above conclusion we can say that The number of moles of 210 grams of NaHCO₃ is 2.5 moles.

Learn more about the Moles here: brainly.com/question/15356425

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In this experiment, the aim is to measure the specific latent heat of fusion of water. In order to do that, a mix of ice/water is heated with a certain amount of heat Q, and then it is evaluated the amount of mass m that undergoes melting.

An important precaution that must be taken in this experiment is that the temperature of the water/ice mixture remains constant and equal to the melting point of ice (otherwise, part of the heat supplied to the mixture is used to increase the temperature of the water, resulting in an inaccurate measure of Q). Therefore, a thermometer must be used to check constantly the temperature, and this thermometer must be placed inside the ice/water mixture only (It should not touch the saucepan, whose temperature can be higher).

So, the correct answer is

in contact with the ice/water only

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3 years ago

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sergeinik [125]

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$C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

$C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole$ ..[1]

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Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, Using Hess's law:

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$\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3$

$\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)$

$\Delta H=-238.7kJ/mole$

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