The answer will be Magnesium. It is a two positive ion and it can charge electrons and protons.
Answer:
C11H25SO4
Explanation:
The total mass of the compound is 253.4 g, so, the mass of each element will be:
C: 52.14% of 253.4 = 0.5214x253.4 = 132.12 g
H: 9.946% of 253.4 = 0.09946x253.4 = 25.20 g
S: 12.66% of 253.4 = 0.1266x253.4 = 32.08 g
O: 25.26% of 253.4 = 0.2526x253.4 = 64.00 g
The molar mass are: C = 12 g/mol, H 1 g/mol, S = 32 g/mol, and O = 16 g/mol
So, to know how much moles will be, just divide the mass calculated above for the molar mass:
C: 132.12/12 = 11 moles
H: 25.20/ 1 = 25 moles
S: 32.08/32 = 1 mol
O: 64.00/16 = 4 moles
So the molecular formula is C11H25SO4
Answer:
CaCl2 (aq) + K2CO3(aq) ---------> CaCO3(s) + 2KCl(aq)
Explanation:
We have the reactants as calcium chloride and potassium carbonate. Recall that we are expecting that the reaction will yield a precipitate. We must keep that in mind as we seek to write its balanced chemical reaction equation.
So we now have;
CaCl2 (aq) + K2CO3(aq) ---------> CaCO3(s) + 2KCl(aq)
Recall that the rule of balancing chemical reaction equation states that the number of atoms of each element on the right side of the reaction equation must be the same as the number of atoms of the same element on the left hand side of the reaction equation.
Equation is as follow,
<span> 4 Na (s) + O</span>₂ <span>(g) → 2Na</span>₂<span>O (s)
According to equation,
91.92 g (4 moles) of Na produces = 123.92 g (2 moles) of Na</span>₂O
So,
17.4 g of Na will produce = X g of Na₂O
Solving for X,
X = (17.4 g × 123.92 g) ÷ 91.92 g
X = 23.45 g of Na₂O
Answer:
ºC
Explanation:
We have to start with the variables of the problem:
Mass of water = 60 g
Mass of gold = 13.5 g
Initial temperature of water= 19 ºC
Final temperature of water= 20 ºC
<u>Initial temperature of gold= Unknow</u>
Final temperature of gold= 20 ºC
Specific heat of gold = 0.13J/gºC
Specific heat of water = 4.186 J/g°C
Now if we remember the <u>heat equation</u>:
We can relate these equations if we take into account that <u>all heat of gold is transfer to the water</u>, so:
Now we can <u>put the values into the equation</u>:
Now we can <u>solve for the initial temperature of gold</u>, so:
ºC
I hope it helps!
