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2 years ago

14

A 30.5-g sample of water at 300. K is mixed with 48.5 g water at 350. K. Calculate the final temperature of the mixture assuming

no heat loss to the surroundings? I need help solving.

1 answer:

ahrayia [7]2 years ago

8 0

Answer:

sorry but i cant

Explanation:

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Which of the following statements are TRUE for BOTH oxygen and sulfur atoms?

irina1246 [14]

Answer:

b. have the same kind number of complete shells

4 0

2 years ago

A 30-g car rolls from a hill 12 cm high and is traveling at 154 cm/s as it travels along a 275 cm horizontal track. What is the

Olegator [25]

Answer:

Explanation:

By definition, the momentum is given by:

p=m*v

Where,

m: car mass

v: speed of the car

Substituting the values in the given definition we have:

p=3*154

Answer:

the momentum of the car is:

4620g * cm/s

8 0

2 years ago

The temperature in a town started out at 55 degrees. By the end of the day the temperature dropped to -6 degrees.

garik1379 [7]

Answer:

The choice of the answer is fourth option that is -61 degrees.

Therefore the temperature drop is -61°Centigrade.

Explanation:

Given:

The temperature in a town started out at 55 degrees

Start temperature = 55°Centigrade. (Initial temperature)

End of the Day = -6°Centigrade. (Final temperature)

To Find:

How far did the temperature drop?

Solution:

We will have,

$\textrm{Temperature drop}=\textrm{Final temperature}-\textrm{Initial temperature}$

Substituting the above values in it we get

$\textrm{Temperature drop}=-6-55\\\\\therefore \textrm{Temperature drop}=-61\° centigrade$

Therefore the temperature drop is -61°Centigrade.

4 0

3 years ago

18. What is the percent by mass sugar of a solution made by dissolving 12.45 grams of sugar in 100 grams of water?

Fantom [35]

The percent by mass sugar of a solution : 11.07%

<h3>Further explanation</h3>

Given

mass of sugar = 12.45 g

mass of water = 100 g

Required

The percent by mass

Solution

Mass of solution :

$\tt mass~sugar+mass~water=12.45+100=112.45~g$

Percent mass of sugar :

$\tt \%mass=\dfrac{mass~sugar}{mass~solution}\times 100\%\\\\\%mass=\dfrac{12.45}{112.45}\times 100\%=11.07\%$

6 0

2 years ago

Bromine, a liquid at room temperature has a boiling point of 58 degrees celsius and a melting point at -7.2 degree celsius bromi

ra1l [238]

Hello!

Bromine can be classified as a pure substance.

Why?

Bromine is an element with atomic number 35 on group 17 of the Periodic Table. That's the first sign that shows us that it is a pure substance.

But the fact that it has a clear and defined boiling and melting point is a sign that we are in the presence of a pure substance. Pure substances are characterized by defined boiling and melting points.

Mixtures usually have a range of temperatures in which they melt and boil.

Have a nice day!

8 0

3 years ago