Answer:
32.83 grams of NH₄F can be formed from 34.6 grams of CaF₂
Explanation:
The balanced reaction is:
CaF₂ + (NH₄)₂O ⇒ 2 NH₄F + CaO
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following quantities participate in the reaction:
- CaF₂: 1 mole
- (NH₄)₂O: 1 mole
- NH₄F: 2 mole
- CaO: 1 mole
Being the molar mass of the compounds:
- CaF₂: 78 g/mole
- (NH₄)₂O: 52 g/mole
- NH₄F: 37 g/mole
- CaO: 56 g/mole
Then by stoichiometry, the following amounts of mass participate in the reaction:
- CaF₂: 1 mole* 78 g/mole= 78 g
- (NH₄)₂O: 1 mole* 52 g/mole= 52 g
- NH₄F: 2 mole* 37 g/mole= 74 g
- CaO: 1 mole* 56 g/mole= 56 g
You can apply the following rule of three: if 78 grams of CaF₂ form 74 grams of NH₄F by stoichiometry, 34.6 grams of CaF₂ will form how much mass of NH₄F?
mass of NH₄F= 32.83 grams
<u><em>32.83 grams of NH₄F can be formed from 34.6 grams of CaF₂</em></u>
<u><em></em></u>
Answer: D
Explanation:
The sun shines the sputhern hemisphere fully on the top= summer
the sun shines somewhat towards the Southern Hemisphere but itfaces away= fall
sun doesnt show at the bottom= winter
faces it more than example b= D
Answer: The enthalpy of formation of 
is -396 kJ/mol
Explanation:
Calculating the enthalpy of formation of
The chemical equation for the combustion of propane follows:
The equation for the enthalpy change of the above reaction is:
![\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_2%28g%29%29%7D%29%2B%281%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28O_2%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![-198=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta -297)+(1\times (0))]\\\\\Delta H^o_f_{(SO_3(g))}=-396kJ/mol](https://tex.z-dn.net/?f=-198%3D%5B%282%5Ctimes%20%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20-297%29%2B%281%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H%5Eo_f_%7B%28SO_3%28g%29%29%7D%3D-396kJ%2Fmol)
The enthalpy of formation of is -396 kJ/mol
The universe comes into existence is first
The first neutral atoms form is second
The universe begins expanding is third
Gases form that will later go to shape stars and galaxies is fourth
Atomic nuclei form is last
I'm almost certain that is correct. Do not take my word for this.
Answer:
1. Molecular equation
BaCl2(aq) + 2AgNO3(aq) –> 2AgCl(s) + Ba(NO3)2 (aq)
2. Complete Ionic equation
Ba²⁺(aq) + 2Cl¯(aq) + 2Ag⁺(aq) + 2NO3¯ (aq) —> 2AgCl(s) + Ba²⁺(aq) + 2NO3¯(aq)
3. Net ionic equation
Cl¯(aq) + Ag⁺(aq) —> AgCl(s)
Explanation:
