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sveta [45]
3 years ago
11

If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is Group of answer choices

Chemistry
1 answer:
BaLLatris [955]3 years ago
3 0

Answer:

The water is completely vaporized at this stage.

Explanation:

The complete question is

If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is

-boiling

-completely vaporized

-frozen solid

-decomposed

-still a liquid

Energy added = 50 kJ = 50000 J

mass of water = 15.5 g = 0.0155 kg

temperature of water = 10 °C

We know that the energy posses by a mass of water at a given temperature is given as

H = mcT

where H is the energy possessed by the mass of water

m is the mass of the water

c is the specific heat capacity of water = 4200 J/ kg- °C

T is the temperature of the water

substituting values, the energy of this amount of water is

H = 0.0155 x 4200 x 10 = 651 J

If 50 kJ is added to the water, the energy increases to

50000 J + 651 J = 50651 J

Temperature of this water at this stage will be gotten from

H = mcT

we solve for the new temperature

50651 = 0.0155 x 4200 x T

50651 = 65.1 x T

T = 50651/65.1 = 778.05 °C

This temperature is well over 100 °C, which is the vaporization temperature of water, but less than 3000 °C for its molecules to decompose.

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A sailor on a trans-Pacific solo voyage notices one day that if he puts 375. mL of fresh water into a plastic cup fresh water we
ElenaW [278]

Answer:

Amount of salt dissolved in each liter of seawater = 40 g

Explanation:

According to Archimedes principle, a body will float in a fluid if the upthrust experienced by a body is equal to the to the weight of the body.

Also, the volume of seawater displaced equals the volume of freshwater in the cup.

From the above principle, since the freshwater and cup floats in the seawater, their combined weight equals the upthrust.

Therefore, mass of equal volume of displaced seawater = mass of freshwater + mass of cup

Mass of freshwater = density of freshwater * volume

density of freshwater = 1 g/mL; volume = 375 mL

mass of freshwater = 375 mL * 1 g/mL = 375 g

mass of seawater = 375 + 15 = 390 g

mass of salt in 375 mL seawater = mass of seawater - mass of freshwater

mass of salt = (390 - 375) g = 15 g

Since 15 g of salt are dissolved in 375 mL seawater, mass of salt in 1 L of seawater =(1000 mL/ 375) * 15g = 40 g

Therefore, amount of salt dissolved in each liter of seawater = 40 g

3 0
3 years ago
What happens to the electrons in an ionic bond?
Anton [14]

Answer:

In an ionic bonds, the metal loses electrons to become a positively charged cation, In which the nonmetal accepts those electrons to become a negatively charged anion.

Explanation:

3 0
3 years ago
Read 2 more answers
Calculate the pressure of 2.50 Liters of a gas at 25.0oC if it has a volume of 4.50 Liters at
MariettaO [177]

Answer:

P_2 =0.51  atm

Explanation:

Given that:

Volume (V1) = 2.50 L

Temperature (T1) = 298 K

Volume (V2) = 4.50 L

at standard temperature and pressure;

Pressure (P1) = 1 atm

Temperature (T2) = 273 K

Pressure P2 = ??

Using combined gas law:

\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \\ \\ \dfrac{1 *2.5}{298} = \dfrac{P_2*4.5}{273}

0.008389261745 \times 273 = 4.5P_2

P_2 =\dfrac{0.008389261745 \times 273 }{4.5}

P_2 =0.51 \ atm

4 0
2 years ago
Consider the following reaction:
lesya692 [45]

Answer:

A

Explanation:

I think it's the answer

6 0
2 years ago
51.7ml at 27 Celsius and 90kpa to stp
never [62]
STP is the abbreviation of standard condition for temperature and pressure which is 273.15K temperature and 1.013× 10^5 Pa pressure. Since the pressure and temperature changes, I assume the question would ask about the result of the volume. The temperature used in ideal gas should be Kelvin, so 27 Celcius would be 300.15K.
The calculation would be
PV=T
V=T/P

V2/V1= T2*P1/T1*P2
V2/V1=273.15K*  90^10^3Pa/ 300.15K *  1.013× 10^5 Pa
V2= 0.81904 * 51.7ml
V2= 42.34ml
6 0
3 years ago
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