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Kitty [74]
3 years ago
14

How much eat is absorbed when 50.0g of water changes from 25.0 Celsius to 37.0 Celsius?

Chemistry
2 answers:
Katyanochek1 [597]3 years ago
6 0
4.186 j/g. X 50g x (37-25) =
2,511 joules or 2.5 kilo joules
Or 2.38 btu


4.186 is specific heat of water per gram per deg C at sea level
Times mass of 50 grams
Times delta T of the water


2530 joules when cross checked by plugging into an online calculator

Firdavs [7]3 years ago
5 0
27 I think beacuse if you subtract
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HELP PLEASE THIS IS CONFUSING
Rasek [7]

Number of moles = 8.3 x 10⁻¹³

<h3>Further explanation </h3>

The mole is the number of particles(molecules, atoms, ions) contained in a substance

1 mol = 6.02.10²³  particles

Can be formulated

N=n x No

N = number of particles

n = mol

No = Avogadro's =  6.02.10²³

5 x 10¹¹ atoms of Silver :

\tt mol=\dfrac{5\times 10^{11}}{6.02\times 10^{23}}=8.3\times 10^{-13}

5 0
3 years ago
Which of the following is an example of a pure substance? *
Natali [406]

Answer:

Iron shavings

Explanation:

A substance that has a fixed chemical composition throughout and a mixture is when two or more substances are combined, but they are not combined chemically.

3 0
3 years ago
Fatima is designing an advanced graphics card for a next-generation video game system. She needs to be able to precisely control
Salsk061 [2.6K]

Answer:svbehsbss c

Explanation:

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8 0
3 years ago
How much heat is gained when a 50.32g piece of aluminum is heated from 9.0°c to 16°c
Rashid [163]

Answer: 317 joules

Explanation:

The quantity of heat energy (Q) gained by aluminium depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

In this case,

Q = ?

Mass of aluminium = 50.32g

C = 0.90J/g°C

Φ = (Final temperature - Initial temperature)

= 16°C - 9°C = 7°C

Then, Q = MCΦ

Q = 50.32g x 0.90J/g°C x 7°C

Q = 317 joules

Thus, 317 joules of heat is gained.

5 0
4 years ago
Read 2 more answers
The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure
Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate \Delta H_{vap} of the reaction, we use clausius claypron equation, which is:

\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

P_1 = vapor pressure at temperature T_1

P_2 = vapor pressure at temperature T_2

\Delta H_{vap} = Enthalpy of vaporization  

R = Gas constant = 8.314 J/mol K

1) \Delta H_{vap}=40.68 kJ/mol=40680 J/mol

T_1 = initial temperature =363 K

T_2 = final temperature =373 K

P_2=1 atm, P_1=?

Putting values in above equation, we get:

\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]

P_1=0.69671 atm \approx 0.6970 atm

The vapor pressure at temperature 363 K is 0.6970 atm

2) \Delta H_{vap}=40.68 kJ/mol=40680 J/mol

T_1 = initial temperature =373 K

T_2 = final temperature =383 K

P_1=1 atm, P_2?

Putting values in above equation, we get:

\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]

P_2=1.4084 atm \approx 1.410 atm

The vapor pressure at 383 K is 1.410 atm

8 0
3 years ago
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