Answer:
Approx. 8 grams of KNO3 will not dissolve
Explanation:
According to the curve at 20 degrees C only 32 grams of KNO3 can dissolve in 100 grams of water meaning if you hvae 40 grams of KNO3 in water at 20 degrees C ; 40-32= 8
Answer:
The answer is A
Explanation:
Because . . .
- A is correct because scientist make their statements by using their observations and tools to measure.
- B is incorrect people's opinions are not facts! Plus, opinions can be wrong.
- C is incorrect because that book can be outdated and not every answer is in a book.
- D is absolutely incorrect because guessing a "right" answer can be wrong. For example, if you were a scientist and you guessed a treatment for coronavirus, you will defiantly get fired because you don't have evidence or logic to back up your answer. In addition, if people used ur "treatment" the one you have guessed as the "right" answer, you can put people's lives in danger and they can possibly die. You will defiantly be responsible for it and be put into jail.
The best way to display this data for analysis would be a line graph
because line graphs are used to track changes over short periods of
time. when smaller changes exist. Line graphs can also be used to
compare changes over the same period of time for more than one group.
Answer:
Explanation:
Substance are consist of very large number of atoms, ions , molecules etc.
They can not be count in a simple way. Scientist used Avogadro number to express the large number of atoms, ions, molecules.
Avogadro number:
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ molecules of water
1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen
96.06 g of SO₄²⁻ ion = 1 mole of SO₄²⁻ ion = 6.022 × 10²³ ions of SO₄²⁻
Answer:
The Standard enthalpy of reaction:
Explanation:
Given- Standard Heat of Formation:
= -904.6 kJ/mol
= 0 kJ/mol,
= +66.4 kJ/mol
= -285.8 kJ/mol
<u><em>Given chemical reaction:</em></u> H₃AsO₄(aq) + 4H₂(g) → AsH₃(g) + 4H₂O(l)
<em>The standard enthalpy of reaction:</em>
= ?
<u><em>To calculate the Standard enthalpy of reaction</em></u> (
)<em><u>, we use the equation:</u></em>

![\Delta H_{r}^{\circ } = [1 \times \Delta H_{f}^{\circ } [AsH_{3} (g)] + 4 \times \Delta H_{f}^{\circ } [H_{2}O(l)]] - [1 \times \Delta H_{f}^{\circ } [H_{3}AsO_{4}(aq)] + 4 \times \Delta H_{f}^{\circ } [H_{2}(g)]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Br%7D%5E%7B%5Ccirc%20%7D%20%3D%20%5B1%20%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B%5Ccirc%20%7D%20%5BAsH_%7B3%7D%20%28g%29%5D%20%2B%204%20%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B%5Ccirc%20%7D%20%5BH_%7B2%7DO%28l%29%5D%5D%20-%20%5B1%20%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B%5Ccirc%20%7D%20%5BH_%7B3%7DAsO_%7B4%7D%28aq%29%5D%20%2B%204%20%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B%5Ccirc%20%7D%20%5BH_%7B2%7D%28g%29%5D)
![\Rightarrow \Delta H_{r}^{\circ } = [1 \times (+66.4\,kJ/mol) + 4 \times (-285.8\,kJ/mol) ] - [1 \times (-904.6\,kJ/mol) + 4 \times (0\,kJ/mol)]](https://tex.z-dn.net/?f=%5CRightarrow%20%5CDelta%20H_%7Br%7D%5E%7B%5Ccirc%20%7D%20%3D%20%5B1%20%5Ctimes%20%28%2B66.4%5C%2CkJ%2Fmol%29%20%2B%204%20%5Ctimes%20%28-285.8%5C%2CkJ%2Fmol%29%20%5D%20-%20%5B1%20%5Ctimes%20%28-904.6%5C%2CkJ%2Fmol%29%20%2B%204%20%5Ctimes%20%280%5C%2CkJ%2Fmol%29%5D)
![\Rightarrow \Delta H_{r}^{\circ } = [-1076.8\, kJ] - [-904.6\,kJ]](https://tex.z-dn.net/?f=%5CRightarrow%20%5CDelta%20H_%7Br%7D%5E%7B%5Ccirc%20%7D%20%3D%20%5B-1076.8%5C%2C%20kJ%5D%20-%20%5B-904.6%5C%2CkJ%5D)

<u>Therefore, the Standard enthalpy of reaction:</u>