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gregori [183]
3 years ago
8

If 747 joules of energy is added to a sample of aluminum (specific heat capacity = 0.899 J/g*C) and the temperature goes up from

105.0*C to 121.0*C, what is the mass of the aluminum sample?
Chemistry
1 answer:
ser-zykov [4K]3 years ago
3 0

Answer:

m = 51.93 grams

Explanation:

Given that,

Energy added, Q = 747 joules

The specific heat capacity of Aluminium, c = 0.899 J/g°C

The temperature goes from 105.0°C to 121.0°C.

Let the mass of the aluminum sample is m. We know that, the heat added to the system is given by :

Q=mc\Delta T\\\\m=\dfrac{Q}{c\Delta T}\\\\m=\dfrac{747}{0.899 \times (121-105)}\\\\m=51.93\ g

So, the mass of the sample is 51.93 grams.

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When an object starts moving on its own,potential energy becomes kinetic energy.?
igomit [66]
Yes because an object has kinetic energy when its moving only, and has only potential energy when velocity is zero and there is height but for your statement to be true,the object has to have some height so there can be potential which can become kinetic energy 
7 0
3 years ago
This is due today so please help me
Nonamiya [84]
B north to south. Hope this helps
7 0
3 years ago
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A scientist fills a large, tightly sealed balloon with 75,000 mL of helium at STP.
pav-90 [236]

Considering the definition of STP conditions, 3.35 moles of helium are contained within the balloon.

<h3>Definition of STP condition</h3>

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

<h3>Amount of moles of helium within the balloon</h3>

In this case, you know that scientist fills a large, tightly sealed balloon with 75,000 mL of helium at STP.

So, you can apply the following rule of three: if by definition of STP conditions 22.4 liters are occupied by 1 mole of helium, 75 L (75 L= 75000 mL, being 1 L= 1000 L) are occupied by how many moles of helium?

amount of moles of helium=\frac{75 Lx1 mole}{22.4 L}

<u><em>amount of moles of helium= 3.35 moles</em></u>

Finally, 3.35 moles of helium are contained within the balloon.

Learn more about STP conditions:

brainly.com/question/26364483

brainly.com/question/8846039

brainly.com/question/1186356

3 0
3 years ago
How many grams of Na2SO4 should be weighed out to prepare 0.5L of a 0.100M solution?​
Nezavi [6.7K]

Answer:

7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100 M solution.

Explanation:

First of all the molecular weight of Na2SO4 is 142.08 gram.Now we all know that if the molecular weight of a compound is dissolved in 1000ml or 1 litee of water then the strength of that solution becomes 1 M.

    According to the given question we have to prepare 0.100 M solution

1000 ml of solution contain 142.08×0.1= 14.208 gram Na2SO4

1    ml of solution contain     14.208÷1000= 0.014 gram

0.5L or 500ml of solution contain 0.014×500= 7gram Na2SO4.

 So it can be stated that 7 gram of Na2SO4 should be required to prepare 0.5L of a 0.100M solution.

     

3 0
3 years ago
1,4-Pentadiene has a AHhydro = -254 kJ/mol while trans-1,3-pentadiene has a AHhydra = -226 kJ/mol. Explain this difference in he
julsineya [31]

Answer:

trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.

Explanation:

Here, \Delta H_{hydro}=H(hydrogenated pdt.)-H(diene)

H(hydrogenated pdt.) is same for both 1,4-pentadiene and 1,3-pentadiene as they both produce pentane after hydrogenation

H(diene) depends on stability of diene.

More stable a diene, lesser will be it's H(diene) value (more neagtive).

trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.

Hence, \Delta H_{hydro} is higher (less negative) for trans-1,3-pentadiene

5 0
4 years ago
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