Answer:

Explanation:
= Mass of metal = 19 g
= Specific heat of the metal
= Temperature difference of the metal = 
V = Volume of water = 150 mL = 
= Density of water = 
= Specific heat of the water = 4.186 J/g°C
= Temperature difference of the water = 
Mass of water

Heat lost will be equal to the heat gained so we get

The specific heat of the metal is
.
Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen
if it occupies 31.6 mL at
.
Explanation:
Given : Mass of oxygen = 0.023 g
Volume = 31.6 mL
Convert mL into L as follows.

Temperature = 
As molar mass of
is 32 g/mol. Hence, the number of moles of
are calculated as follows.

Using the ideal gas equation calculate the pressure exerted by given gas as follows.
PV = nRT
where,
P = pressure
V = volume
n = number of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the value into above formula as follows.

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen
if it occupies 31.6 mL at
.
Answer:
If an object is moving at a constant speed in a constant rightward direction, then the acceleration is zero and the net force must be zero.
The answer to your question is B.
Answer:
1) Oil is less dense than water so when oil spills, it spreads across the entire water surface.
2) The oil spreads very quickly with lighter oils such as gasoline.
3) Wind, Currents, and Warm Temperatures will cause Oil to spread quicker.