The amount of heat lost by granite is equal to the amount
of heat gained by water. Therefore their change in enthalpies must be equal.
The opposite in sign means that one is gaining while the other is losing
ΔH granite = - ΔH water
ΔH is the change in enthalpy experienced by a closed object
as it undergoes change in energy. This is expressed mathematically as,
ΔH = m Cp (T2 – T1)
Given this information, we can say that:
12.5 g * 0.790 J / g ˚C * (T2 – 82 ˚C) =
- 25.0 g * 4.18 J / g ˚C
* (T2 – 22 ˚C)
9.875 (T2 – 82) = 104.5 (22 – T2)
9.875 T2 – 809.75 = 2299 – 104.5 T2
114.375 T2 = 3108.75
T2 = 27.18 ˚C
The temperature of 2 objects after reaching thermal
equilibrium is 27.18 ˚<span>C.</span>
If this molecule is one half of a buffer, then the formula of the second half of the buffer is M2CrO4 where M is a univalent metal.
<h3>What is a strong acid?</h3>
A weak acid is one that is able to ionize completely in solution. The acid called chromic acid H2CrO4 is not able to ionize completely in solution.
We know that a buffer is composed of a weak acid and its salt or a weak base and its salt hence if the acid H2CrO4 is present in a buffer then the other half must be salt of the acid.
If this molecule is one half of a buffer, then the formula of the second half of the buffer is M2CrO4 where M is a univalent metal.
Learn more about buffer:brainly.com/question/22821585
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Answer:
simple
Explanation:
The glow stick's outer plastic tube holds a solution of an oxalate ester and an electron-rich dye along with a glass vial filled with a hydrogen peroxide solution. ... Glow sticks light up when oxalate esters react with hydrogen peroxide to form a high-energy intermediate
Answer:
In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.
Explanation:
tally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:
\displaystyle \%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%%H=
mass compound
mass H
×100%
\displaystyle \%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%%C=
mass compound
mass C
×100%
If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:
\displaystyle \%\text{H}=\frac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%%H=
10.0g compound
2.5g H
×100%=25%
\displaystyle \%\text{C}=\frac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%%C=
10.0g compound
7.5g C
×100%=75%
