Electrons orbit the nucleus in specific, defined paths. Each path has a specified energy.

How many Joules are released to cool 250.0 grams of liquid water from 100°C to 0°C? The specific heat of water is 4.180 J/g.C.

nlexa [21]

$\bold{\huge{\orange{\underline{ Solution}}}}$

$\bold{\underline{ Given :- }}$

We have 250g of liquid water and it needs to be cool at temperature from 100° C to 0° C
Specific heat of water is 4.180J/g°C

$\bold{\underline{ To \: Find :- }}$

We have to find the total number of joules released.

$\bold{\underline{ Let's \:Begin:- }}$

We know that,

Amount of heat energy = mass * specific heat * change in temperature

That is, 

$\sf{\red{ Q = mcΔT }}$

Subsitute the required values in the above formula :-

$\sf{ Q = 250 × 4.180 ×(0 - 100 )}$

$\sf{ Q = 250 × 4.180 × - 100 }$

$\sf{ Q = 250 × - 418}$

$\sf{\pink{ Q = - 104,500 J }}$

Hence, 104,500 J of heat is released to cool 250 grams of liquid water from 100° C to 0° C.

$\bold{\underline{ Now :- }}$

We have to tell whether the above process is endothermic or exothermic :-

Here, In the above process ΔT is negative and as a result of it Q is also negative that means above process is Exothermic

Exothermic process :- It is the process in which heat is evolved .
Endothermic process :- It is the process in which heat is absorbed .

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valentinak56 [21]

Answer: The number of grams of $H_2$ in 1620 mL is 1.44 g

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 1 atm (at STP)

V = Volume of gas = 1620 ml = 1.62 L (1L=1000ml)

n = number of moles = ?

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $273K$

$n=\frac{PV}{RT}$

$n=\frac{1atm\times 16.2L}{0.0821Latm/K mol\times 273K}=0.72moles$

Mass of hydrogen = $moles\times {\text {Molar mass}}=0.72mol\times 2g/mol=1.44g$

The number of grams of $H_2$ in 1620 mL is 1.44 g

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Answer:

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Which best describes the role of a primary consumer in a food web? (3 points)

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Answer:

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Explanation:

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