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Black_prince [1.1K]
2 years ago
12

HURRY PLEASE HELP !!!

Chemistry
1 answer:
melomori [17]2 years ago
5 0

Answer:

I ONLY KNOW ONE THAT IS

2C2H6+5O2=>4CO+6H2O

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(Please help will mark brainliest!)
arsen [322]

We know that Weighted atomic mass of Element is Calculated based upon their existence of isotopes and the Relative abundance of these isotopes.

Given that the Element X is Composed of :

Isotope                       Relative abundance                   Atomic Mass

⁵⁵X                                        70%                                               55

⁵⁶X                                        20%                                               56

⁵⁷X                                        10%                                                57

Weighted Atomic Mass of Element X :

= (0.70)(55) + (0.20)(56) + (0.10)(57)

= 38.50 + 11.2 + 5.70

= 55.4

So, the Weighted Atomic Mass of Element X is 55.4

6 0
2 years ago
Calculate the molarity of a solution of sodium hydroxide, naoh, if 23.64 ml of this solution is needed to neutralize 0.5632 g of
GuDViN [60]
The  molarity  of  NaOH  needed  is  calculated   as  follows
calculate  the  moles  of  KhC8h4O4

that  is  moles  =  mass/molar  mass  of  KhC8h4O4(204.22 g/mol)

=0.5632g /204.22g/mol=  2.76  x10^-3  moles

write the  equation  for  reaction

khc8h4O4  +  NaOH  ---> KNaC8h4O4  +  H2O

from  the  equation  above  the   reacting  ratio   of   KhC8h4O4  to  NaOh  is  1:1  therefore  the  moles  of  Naoh  is  also  2.76  x10^-3  moles

molarity  of NaOh  =  (moles  of  NaOh /  volume ) x  1000

that  is { (2.76  x10^-3) / 23.64}  x100  =0.117 M
8 0
3 years ago
How do you toast a toaster?
Mama L [17]

by putting to much current through it ?

7 0
3 years ago
During oxidation, which of the following is true? A. Electrons are gained, so the oxidation number increases. B. Electrons are g
EastWind [94]

Oxidation is when the overall charge (or oxidation number) increases.  The only way to increase an oxidation number is to lose an electron, thereby making the negative charges less.  The correct answer is C.

6 0
3 years ago
Read 2 more answers
How do I calculate the standard enthalpy change for the following reaction at 25 °C ?
ladessa [460]
MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g) 

Using the standard enthalpies of formation given in the source below: 

(−601.24 kJ) + (2 x −92.30 kJ) − (−641.8 kJ) − (−285.8 kJ) = +141.76 kJ 
So: 

MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g), ΔH = +141.76 kJ
4 0
3 years ago
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