and
.
Assuming complete decomposition of both samples,
First compound:
;
of the first compound would contain
Oxygen and mercury atoms seemingly exist in the first compound at a
ratio; thus the empirical formula for this compound would be
where the subscript "1" is omitted.
Similarly, for the second compound
;
of the first compound would contain
and therefore the empirical formula
.
Answer:
2
Explanation:
Carbon Is tetravalent. This means that at any particular point in time, the number of bonds carbon can form at a particular time is 4. Now in this particular question, there is already a double bond between the two carbon atoms. This means that each of the carbon atom has the chance to fulfill it tetra valency by attaching just two bonds to itself.
Hence, to complete the property of its tetra valency, two more bonds needed to be added to the two carbon atoms
Answer:
B.) If sodium carbonate is added to vinegar, the reaction will absorb heat.
Explanation:
A.) is incorrect because this is not testable. Rather, it is just an opinion that cannot be proven correct or incorrect.
B.) is correct because this statement is testable. Tests need to be run to determine the accuracy of the statement.
C.) is incorrect because this statement explains something that does not need to be tested. It is an example of a physical change when one tears a piece of paper.
D.) is incorrect because this is already a true statement. It is obvious that not all reactions absorb/release heat. While tests could be run to further prove this statement true, it is already considered accurate.
4% mass / volume :
4 g ---------> 100 mL
1.2 g ------- ? mL
V = 1.2 * 100 / 4
V = 120 / 4
V = 30 mL
hope this helps!
Answer:
1.71x10²⁷
Explanation:
If we sum 1/2 of (3) + 1/2 of (1):
1/2 (3.) C(s) + 1/2O₂(g) ⇌ CO(g), K₃ = √2.10×10⁴⁷ = 4.58x10²³
1/2 (1) 1/2CO₂(g) + 3/2H₂(g) ⇌ 1/2CH₃OH(g) + 1/2H₂O(g), K₁ = √1.40×10² = 11.8
C(s) + 1/2O₂(g) +<u> 1/2CO₂(g) </u>+<u> 3/2H₂(g</u>) ⇌ 1/2CH₃OH(g) + <u>1/2H₂O(g)</u> + <u>CO(g)</u>
K' = 4.58x10²³ * 11.8 = 5.42x10²⁴
+1/2 (2):
<u>1/2 CO(g)</u> +<u> 1/2H₂O(g)</u> ⇌<u> 1/2CO₂(g)</u> + <u>1/2H₂</u> (g), K = √1.00×10⁵ = 316.2
C(s) + 1/2O₂(g) + H₂(g) ⇌ 1/2 CHO₃H(g) + 1/2CO(g)
K'' = 5.42x10²⁴* 316.2 =
<h3>1.71x10²⁷</h3>