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3 years ago

Define a force.......

Chemistry

2 answers:

FinnZ [79.3K]3 years ago

6 0

Answer:

Gravitational Force.

Gravitation is the agent that gives weight to objects with mass and causes them to fall on the ground when dropped.

grin007 [14]3 years ago

5 0

Answer:

A force is the strength or energy as an attribute of physical action or movement.

Explanation:

In smaller words, a force is a form of energy caused from a push or a pull on an object. A force happens when two objects interact with each other. For example, when you throw a ball you are using force.

Hopefully this helps im sorry if i got it wrong im kinda dumb... XD

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What is the value for AG at 100 Kif AH = 27 kJ/mol and AS = 0.09 kJ/(mol-K)?

Elina [12.6K]

Answer:

ΔG = 18KJ/mol

Explanation:

Given data:

ΔS = 0.09 Kj/mol.K

ΔH = 27 KJ/mol

Temperature = 100 K

ΔG = ?

Solution:

Formula:

ΔG = ΔH - TΔS

ΔH = enthalpy

ΔS = entropy

by putting values,

ΔG = 27 KJ/mol - 100K(0.09 Kj/mol.K)

ΔG = 27 KJ/mol - 9 KJ/mol

ΔG = 18KJ/mol

7 0

2 years ago

The term element refers to atoms that have the same number of _______.

erica [24]

I guess it’s Electrons

7 0

3 years ago

Read 2 more answers

Find the number of moles of argon in 364 grams of argon.

serg [7]

Answer:D. 9.1 mole Ar

Explanation:

364 g Ar x 1 mole Ar / 40 g Ar

= 9.1 moles Ar

3 0

3 years ago

A 0.150-kg sample of a metal alloy is heated at 540 Celsius an then plunged into a 0.400-kg of water at 10.0 Celsius, which is c

Zarrin [17]

Answer:

$C_{alloy}=0.497\frac{J}{g\°C}$

Explanation:

Hello there!

In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:

$Q_{allow}=-(Q_{water}+Q_{Al})$

Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:

$m_{alloy}C_{alloy}(T_{eq}-T_{alloy})=-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})$

Then, we solve for specific heat of the metallic alloy to obtain:

$C_{alloy}=\frac{-(m_{water}C_{water}(T_{eq}-T_{water})+m_{Al}C_{Al}(T_{eq}-T_{Al})}{m_{alloy}(T_{eq}-T_{alloy})}$

Thereby, we plug in the given data to obtain:

$C_{alloy}=\frac{-(400g*4.184\frac{J}{g\°C} (30.5\°C-10.0\°C)+200g*0.900\frac{J}{g\°C}(30.5\°C-10.0\°C)}{150g(30.5\°C-540\°C)} \\\\C_{alloy}=0.497\frac{J}{g\°C}$

Regards!

3 0

3 years ago

Sea water contains roughly 158.0g of Nacl per liter. What is the molarity of sodium chloride in sea water?

IrinaK [193]

Molar mass NaCl = 58.5 g/mol

C = 158.0 g/L

Molarity = C / molar mass

M = 158.0 / 58.5

M = 2.7000 M

hope this helps!

6 0

3 years ago