Answer:
We need 8.11 grams of glucose for this solution
Explanation:
Step 1: Data given
Molarity of the glucose solution = 0.300 M
Total volume = 0.150 L
The molecular weight of glucose = 180.16 g/mol
Step 2: Calculate moles of glucose in the solution
Moles glucose = molarity solution * volume
Moles glucose = 0.300 M * 0.150 L
Moles glucose = 0.045 moles glucose
Step 3: Calculate mass of glucose
MAss glucose = moles glucose* molecular weight of glucose
MAss glucose = 0.045 moles * 180.16 g/mol
MAss glucose = 8.11 grams
We need 8.11 grams of glucose for this solution
Answer: 0.5 mole Mg
Explanation: solution:
12 g Mg x 1 mole Mg / 24 g Mg
= 0.5 mole Mg
Answer:
See below
Step-by-step explanation:
- Hydrogen either reacts with or is formed by reactions with many other elements, so chemists could use it directly to determine their relative masses.
- Hydrogen has the smallest atomic mass, so it was convenient to give H a relative atomic mass of 1 and assign those of other elements as multiples of this number.
The O = 16 scale became the standard in 1903 and carbon-12 was chosen in 1961.
The common substance among the product(s) of the first equation and among the reactant(s) in the second equation is H2O(g). We can eliminate that as an intermediate. The overall chemical equation will thus be:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l),
which is the first answer choice.
In essence, all you’re doing here is swapping water vapor for liquid water.
Answer:
41.17g
Explanation:
We are given the following parameters for Flourine gas(F2).
Volume = 5.00L
Pressure = 4.00× 10³mmHG
Temperature =23°c
The formula we would be applying is Ideal gas law
PV = nRT
Step 1
We find the number of moles of Flourine gas present.
T = 23°C
Converting to Kelvin
= °C + 273k
= 23°C + 273k
= 296k
V = Volume = 5.00L
R = 0.08206L.atm/mol.K
P = Pressure (in atm)
In the question, the pressure is given as 4.00 × 10³mmHg
Converting to atm(atmosphere)
1 mmHg = 0.00131579atm
4.00 × 10³ =
Cross Multiply
4.00 × 10³ × 0.00131579atm
= 5.263159 atm
The formula for number of moles =
n = PV/RT
n = 5.263159 atm × 5.00L/0.08206L.atm/mol.K × 296K
n = 1.0834112811moles
Step 2
We calculate the mass of Flourine gas
The molar mass of Flourine gas =
F2 = 19 × 2
= 38 g/mol
Mass of Flourine gas = Molar mass of Flourine gas × No of moles
Mass = 38g/mol × 1.0834112811moles
41.169628682grams
Approximately = 41.17 grams.