Answer:
what are the roles of minerals in maintaning homeostasis?
Answer:

Explanation:
This question asks us to find the temperature change given a volume change. We will use Charles's Law, which states the volume of a gas is directly proportional to the temperature. The formula is:

The volume of the gas starts at 250 milliliters and the temperature is 137 °C.

The volume of the gas is increased to 425 milliliters, but the temperature is unknown.

We are solving for the new temperature, so we must isolate the variable T₂. First, cross multiply. Multiply the first numerator and second denominator, then multiply the first denominator and second numerator.

Now the variable is being multiplied by 250 milliliters. The inverse of multiplication is division. Divide both sides of the equation by 250 mL.


The units of milliliters (mL) cancel.



The temperature changes to <u>232.9 degrees Celsius.</u>
The tea was no longer hot or (brewed) so the 5th didn’t dissolve like the others because the tea was hot or warm enough anymore it cooled down. So the sugar won’t dissolve no more.
Answer:
Strong acid breaks up into ions
Explanation:
Answer:
True => ΔH°f for C₆H₆ = 49 Kj/mole
Explanation:
See Thermodynamic Properties Table in appendix of most college level general chemistry texts. The values shown are for the standard heat of formation of substances at 25°C. The Standard Heat of Formation of a substance - by definition - is the amount of heat energy gained or lost on formation of the substance from its basic elements in their standard state. C₆H₆(l) is formed from Carbon and Hydrogen in their basic standard states. All elements in their basic standard states have ΔH°f values equal to zero Kj/mole.