Answer:
5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O=
Answer: 9.04 g of H2O
Explanation:
First set up equation: C4H10 (g)+ O2(g) -> CO2(g) + H2O(g)
Next balance it: 2C4H10 (g)+ 13O2(g) -> 8CO2(g) + 10H2O (g)
Use equation to get moles and plug given
5.83g C4H10 x (1 mol C4H10/58.05 g (molar mass of C4H10) x (10 mol H2O/ 2 mol C4H10) x (18.002 g H2O (molar mass of H2O)/ 1 mol H2O
Answer:
Like any wave, a sound wave doesn't just stop when it reaches the end of the medium or when it encounters an obstacle in its path. Rather, a sound wave will undergo certain behaviors when it encounters the end of the medium or an obstacle. Possible behaviors include reflection off the obstacle, diffraction around the obstacle, and transmission (accompanied by refraction) into the obstacle or new medium
Sodium has a lower ionization energy than magnesium describes why sodium reacts vigorously than magnesium chloride.
<h3>Why is sodium more reactive than magnesium?</h3>
- Sodium is more reactive than magnesium because it has the ability to easily lose electron, hence have lower ionization energy.
- Sodium belong to group one on the periodic table and they are called akali metal while magnesium belong to group two on the periodic table and they are called alkali Earth metal.
- Sodium and magnesium belong to the in the 3rd period. Iin the outermost energy level sodium has one electron but magnesium has 2 electrons. Therefore, there is more attraction abetween the nucleus and electrons in magnesium than that of sodium.
Therefore, sodium is more reactive than magnesium chloride because of lower ionization energy.
For more details on sodium reactivity, check the link below.
brainly.com/question/6837593
<span>Forward & falling. Hope this helps!</span>
1. Determine if the ionic substances can break apart into ions.
- e.g. CaCO3 isn't very soluble, do it can't dissolve and dissociate. If it can't pop apart, no ions.
2. Swap the partners for all the other ions that you can get from step 1. You can skip pairings with the same charge - a + can't get close to another + to react.
3. Use solubility, acid/base, and redox rules to see if anything will happen with the ions in solution.<span />
