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Natalka [10]
3 years ago
14

A sample of water is heated from room temperature to just below the boiling point. The overall change in temperature is 72°C. E

xpress this temperature change in kelvins. (a) 345 K (b) 72K (c) 0K (d) 201 K (e)273K
Chemistry
2 answers:
Jlenok [28]3 years ago
7 0

Answer : The correct option is, (a) 345 K

Explanation :

The conversion used for the temperature from degree Celsius to Kelvin is:

K=273.15+^oC

where,

K = temperature in Kelvin

^oC = temperature in centigrade

As we are given the temperature in degree Celsius is, 72

Now we have to determine the temperature in Kelvin.

K=273.15+^oC

K=273.15+(72^oC)

K=345.15\approx 345

Therefore, the temperature in Kelvin is, 345 K

juin [17]3 years ago
3 0

Answer: Option (a) is the correct answer.

Explanation:

It is known that when temperature is given in degree celsius and we want it to convert into kelvin then we need to add 273 into the given degree celsius.

For example, given temperature is 72^{o}C.

Now, to convert it into degree celsius we need to add 273 into it as follows.

   Temperature in Kelvin = (72 + 273) K

                                         = 345 K

Thus, we can conclude that given temperature in Kelvin is 345 K.

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Liquid state have more heat and more energy and a little more intermolecular space than solid. Their molecules are no longer stacked neatly due to presence of air capsules, but they still touch each other. Their structure is uncertain and can flow, their shape will alter quickly and dramatically with outside forces, but they will essentially remain a single mass unless pushed apart.

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8 0
1 year ago
How many mL of 0.506 M HClare needed to dissolve 9.85 g of BaCO3?
Blababa [14]

Answer:

197mL of 0,506M HCl

Explanation:

The reaction of HCl + BaCO₃ is:

BaCO₃(s) + 2HCl → BaCl₂(aq) + CO₂ + H₂O.

The moles of BaCO₃ in 9,85 g are:

9,85 g of BaCO₃ × \frac{1mol}{197,34 g} = <em>0,0499 moles of BaCO₃</em>

As 1 mol of BaCO₃ reacts with two moles of HCl, for a complete reaction of BaCO₃ to dissolve this compound in water you need:

0,0499 moles of BaCO₃ × \frac{2molHCl}{1molBaCO_{3}} =<em> 0,0998 moles of HCl</em>

If you have a 0,506M HCl, you need to add:

0,0998 moles of HCl× \frac{1L}{0,506moles} = 0,197 L ≡ 197mL

I hope it helps!

8 0
3 years ago
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Answer:

Explanation:  15

3 0
3 years ago
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Answer:

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Basically, all o did was write the equations, balance it and solve for them. Also, at the place I stared, I used simultaneous equation to solve it. Multiplying by 8 and also 3.

It's a pretty straightforward question.

At the final step that's missing, I Did

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5 0
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