Answer:
Standard boiling point
Explanation:
Note that there are 2 major units of pressure except Pa .
- bar
- atm
At 1 atm pressure the boiling temperature is called normal boiling point.
At 1 bar pressure the boiling temperature is called standard boiling point
Answer:
The object A will be having the greater density compared to object B.
Explanation:
It is known that density of any object is defined as the mass of any object occupying a given volume. So the ratio of mass and volume will help to determine the density of any object.
From the above equation, it can be seen that the density of any object is directly proportional to the mass of the object and inversely proportional to the volume occupied by the object.
So in the present context, the mass of objects A and B are same and it is 100 g. Thus, the density of object A and object B will be influenced by their volume. As it is given that the volume of object A is 50 cm3 and object B is 100 cm3, then depending upon the relationship of volume and density, the density of both the objects can be determined. As the object with higher volume will be having lesser density as volume is inversely proportional to density. Thus, in the given case the volume of object B is greater than object A and so the object A will be having greater density compared to object B.
The displacement function is given by : r = bt2i + ct3j meters
The velocity function is the derivative of the displacement:
v = r' = 2bti + 3ct2j meter / second
Now, for <span> the velocity vector make an angle of 45.0∘ with the x- and y-axes, the i and j components have to be equal:
</span>2bt = 3ct2 (Now solve for t)
2bt = t (3ct)
either t = zero (rejected)
or t = 2b / 3c seconds (accepted)
Answer:
0.01932 L
Explanation:
First we <u>convert 105 mM to M</u>:
Next we <u>convert 552 mL to L</u>:
Then we use the following equation:
Where:
We<u> input the given data</u>:
- 3 M * V₁ = 0.105 M * 0.552 L
And <u>solve for V₁</u>:
Answer:
Las bebidas gaseosas como las gaseosas están hechas de un soluto de dióxido de carbono gaseoso en un líquido. La solubilidad del dióxido de carbono en el líquido depende de la presión y la temperatura de la lata de refresco, y también de agitar la lata de refresco que introduce burbujas que permanecen ocultas hasta que se abre la lata antes de que burbujee.
Por lo tanto, dado que la presión en la lata de refresco permanece constante, elevar la temperatura, agitar la lata de refresco o congelar el refresco, lo que aumenta la cantidad de dióxido de carbono en la porción líquida, hará que el refresco forme espuma y se derrame.
Explanation: