For the reaction, the equilibrium constant equation will be:
Kc = [NO₂]² / [N₂O₄]
Kc = (0.5)² / 0.025
Kc = 10
The answer is C.
Option B: 4.3
There are following rules for rounding off numbers:
1. If the rounding digit has digit less than 5 (1,2,3 and 4) to its right, all the digits on right side of the rounding digit will drop out.
2. If the rounding digit has digit equal and greater than 5 (5,6,7,8 and 9) to its right, one is added to the rounding digit.
Jeff has to weigh 4.312 g sample of NaCl. To round off the underline number 4.<u>3</u>12, check for the digit right to it. It has 1, which is less than 5 thus according to the Rule 1, all the digits on right side of the rounding digit will drop out and the number gets rounded off to 4.3.
Therefore, 4.3 is most accurate measurement by Jeff.
Answer:
1
Explanation:
The empirical formula describes the simplest whole number ratio of each type of atom in a compound. To find this formula, you need to (1) convert grams of each element to moles (via their atomic masses) and then (2) find the ratio of each element (by dividing each molar value by the smallest mole value).
(Step 1)
Atomic Mass (Zn): 65.380 g/mol
Atomic Mass (C): 12.011 g/mol
Atomic Mass (O): 15.998 g/mol
10.40 grams Zn 1 mole
------------------------ x ------------------------ = 0.159 moles Zn
65.380 grams
1.92 grams C 1 mole
--------------------- x ----------------------- = 0.160 moles C
12.011 grams
7.68 grams O 1 mole
---------------------- x ------------------------ = 0.480 moles O
15.998 grams
(Step 2)
0.159 moles Zn / 0.159 = 1 atom Zn
0.160 moles C / 0.159 = 1 atom C
0.480 moles O / 0.159 = 3 atom O
The empirical formula is Zn₁C₁O₃. Therefore, the value that should be in the space is 1.
Uhh I'm kinda blind so can you hold it a little straight
Answer:
CH4 +O2 => CO2 + 2 H2O
None were in excess
Explanation:
Equation for the reaction is,
CH4 + O2 =>CO2 +2 H2O
No of moles of CH4 = (20 /1000)/24 =0.02 /24 = 0.00083
No of moles of O2 =20 /24000 = 0.0083
CH4 : O2 = 1:1
THEREFORE
None of the gases were in excess.
