1.65g MgO = 1g Mg
1.65 - 1 = 0.65 g of O in MgO
solve it using proportion:
1g Mg / 0.65g O = x (g) Mg / 16g O
or 1 / 0.65 = x / 16
24.6 g is the answer.
if 1 gram of oxygen requires 1.65 grams of Mg
then 16 grams of oxygen will require 16 ( 1.65) or 26.4 grams.
Answer:
0.11mol/dm³
Explanation:
The reaction expression is given as:
HCl + NaOH → NaCl + H₂O
Volume of acid = 25cm³ = 0.025dm³
Volume of base = 18.4cm³ = 0.0184dm³
Concentration of base = 0.15mol/dm³
Solution:
The concentration of hydrochloric acid = ?
To solve this problem, let us first find the number of moles of the base;
Number of moles = concentration x volume
Number of moles = 0.15mol/dm³ x 0.0184dm³ = 0.00276mol
From the balanced reaction equation;
1 mole of NaOH will combine with 1 mole of HCl
Therefore, 0.00276mol of the base will combine with 0.00276mol of HCl
So;
Concentration of acid = 
= 
= 0.11mol/dm³
Answer:
depends
Explanation:
if there was a picture i could help you
Answer:
Flammability is a material’s ability to burn in the presence of <u><em>oxygen.</em></u>
Explanation:
Flammability can be described as the ability of a substance to get ignited. Flammability will lead to fire or combustion. Some substances are highly flammable like Benzene. Other tend to be just flammable. And there are also compounds which will nor be flammable at all as they won't react with oxygen. Examples of these substances include helium, steel or glass.
The flammability of a substance shall be considered a very important aspect when storing or transporting a substance.
Mg reaction with O₂ gas will produce MgO so the equation will be
2Mg+O₂⇒2MgO. (You have to find the equation in order two figure out the number of moles of O₂ that will react with 1 mole of MgO).
The first step is to find the number of moles of Mg in 4.03g of Mg. You can do this by dividing 4.03g Mg by its molar mass (which is 24.3g/mol) to get 0.1658mol Mg. Then you have to find the number of moles of O₂ that will react with 0.1658mol Mg. To do this you need to use the fact that 1mol O₂ will react with 2mol Mg (this reatio is from the chemical equation) so you have to multiply 0.1658mol Mg by (1mol O₂)/(2mol Mg) to get 0.0829mol O₂. From here you would usually use PV=nRT and solve for V However, the question tells us that we are at STP, that means you can use the fact that 22.4L of gas is 1 mol of gas at STP. Using that information we can find the volume of O₂ gas by mulitlying 0.0829mol O₂ by 22.4L/mol to get 1.857L which equals 1857mL.
therefore, 1857mL of O₂ gas will react with 4.03g of Mg.
I hope this helps. Let me know in the comments if anything is unclear.
