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3 years ago

What physical and social systems are involved in the service industry

Chemistry

1 answer:

const2013 [10]3 years ago

6 0

Answer:

Hey mate.....

Explanation:

This is ur answer........

<em>For example, the communications industry includes all of the physical and social systems that produce books, newspapers, magazines, radio and television broadcasts, billboard advertisements, Internet websites, telephones, and telecommunications, such as the GPS.</em>

Hope it helps!

Brainliest pls!

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How does concentration affect boiling point of a solvent?

nalin [4]

Answer / explanation:

How does concentration affect boiling point of a solvent?

The amount by which the boiling point is raised is directly dependent on the concentration of the solute.

The higher the concentration of a solute, the more it is said to be difficult for the solvent molecules to escape into the gas phase.

However, when a non volatile amount of substance is dissolved in a given solvent, the boiling point of the given solvent increases.

The higher the concentration, the more higher the boiling point of a solvent.

It requires a higher temperature for enough solvent molecules to escape , this the boiling point is raised elevatedly

5 0

3 years ago

Eyeglasses provide as much protection as safety goggles. * true or false

Artyom0805 [142]

False. Eyeglasses do not cover all around the eye however safety goggles do

7 0

3 years ago

Read 2 more answers

2. Which set of symbols represents isotopes of the same element?

MariettaO [177]

Answer:

so you can do the maths your self

Explanation:

isotopes are elements with he same proton number but difference in their neutron number

proton number is also known as the atomic number
mass number is the sum of the proton number and the neutron number
mass number = proton number + neutron number

7 0

3 years ago

Natural gas is stored in a spherical tank at a temperature of 13°C. At a given initial time, the pressure in the tank is 117 kPa

drek231 [11]

Answer:

1. the absolute pressure in the tank before filling = 217 kPa

2. the absolute pressure in the tank after filling = 312 kPa

3. the ratio of the mass after filling M2 to that before filling M1 = 1.44

The correct relation is option c ( $\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }$ )

Explanation:

To find -

1. What is the absolute pressure in the tank before filling?

2. What is the absolute pressure in the tank after filling?

3. What is the ratio of the mass after filling M2 to that before filling M1 for this situation?

As we know that ,

Absolute pressure = Atmospheric pressure + Gage pressure

So,

Before filling the tank :

Given - Atmospheric pressure = 100 kPa , Gage pressure = 117 kPa

⇒Absolute pressure ( p1 ) = 100 + 117 = 217 kPa

Now,

After filling the tank :

Given - Atmospheric pressure = 100 kPa , Gage pressure = 212 kPa

⇒Absolute pressure (p2) = 100 + 212= 312 kPa

Now,

As given, volume is the same before and after filling,

i.e. $V_{1}$ = $V_{2}$

As we know that, P ∝ M

⇒ $\frac{p_{1} }{p_{2} } = \frac{m_{1} }{m_{2} }$

⇒ $\frac{m_{2} }{m_{1} } = \frac{p_{2} }{p_{1} }$

⇒ $\frac{m_{2} }{m_{1} } = \frac{312 }{217 } = 1.4378$ ≈ 1.44

Now, as we know that PV = nRT

As V is constant

⇒ P ∝ MT

⇒ $\frac{P}{T}$ ∝ M

⇒ $\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }$

So, The correct relation is c option.

6 0

3 years ago

A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o

yaroslaw [1]

<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

Mass of the unknown metal sample as 58.932 g
Initial temperature of the metal sample as 101°C
Final temperature of metal is 23.68 °C
Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

Therefore; mass of pure water is 45.2 g
Initial temperature of water = 21°C
Final temperature of water is 23.68 °C
Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

= 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

= 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q = m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

= 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

= 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>

We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.

Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

= 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

8 0

3 years ago