Answer:
B) 16 g
Explanation:
First we <u>convert 4 moles of O₂ into moles of H₂</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 4 mol O₂ *
= 8 mol H₂
Finally we <u>convert 8 moles of H₂ into grams</u>, using <em>its molar mass</em>:
- 8 mol H₂ * 2 g/mol = 16 g
Thus, the correct answer is option B).
There are 3 moles of
<span>per 1 mole of salt and 1 mole of
</span>per mole of salt, the total ionic concentrations must be
of
, and
of
Answer:
The molarity is 2M
Explanation:
First , we calculate the weight of 1 mol of NaCl:
Weight 1mol NaCl= Weight Na + Weight Cl= 23 g+ 35, 5 g= 58, 5 g/mol
58,5 g---1 mol NaCl
233,772 g--------x= (233,772 g x1 mol NaCl)/58,5 g= 4 mol NaCl
<em>A solution molar--> moles of solute in 1 L of solution:</em>
2 L-----4 mol NaCl
1L----x0( 1L x4mol NaCl)/4L =2moles NaCl---> 2 M
Answer:
Number of molecules = 1.8267×10^20
Explanation:
From the question, we can deuced that the gases behave ideally, the we can make use of the ideal gas equation, which is expressed below;
PV = nRT
where
P =pressure
V =volume
n = the number of moles
R is the gas constant equal to 0.0821 L·atm/mol·K
T is the absolute temperature
Given:
P = 6.75 atm;
T = 290.0 k,
; V = 1.07 cm³ = 0.001 L
( 6.75 atm)(0.00107 L) = n(0.0821 L·atm/mol·K)(290K)
n = 3.0335167*10^-4 moles
But there are 6.022×10²³ molecules in 1 mole,
Number of molecules = 1.8267×10^20
