Answer:
[OH⁻] = 3.54 × 10⁻¹⁰ M
Solution:
The relationship between pOH and [OH⁻] is expressed as,
pOH = - log [OH⁻]
Substituting the value of pOH,
9.45 = -log [OH⁻]
Finding [OH⁻],
[OH⁻] = 10⁻⁹·⁴⁵ ∴ 10 = Antilog
[OH⁻] = 3.54 × 10⁻¹⁰ M
In my opinion,
when two distinct forces are exerted in opposite directions, the resultant force is the subtraction of those forces
Therefore, 4 N - 4 N = 0 N
The amount to administer to the child is 2,469 mL.
To convert to kilograms (kg), the child's weight in pounds (lb) is multiplied by 0.45359237: m(child) = 72.6 · 0.045359237 = 32.93 kg.
To find m(Medrol), the child's mass in kilograms is multiplied by 1.5 mg/kg.
Thus, m(Medrol) = 32.93 kg · 1.5 mg/kg = 49.39 mg.
The concentration of Medrol is d(Medrol) = 20.0 mg/mL.
To find the volume of Medrol needed, use V(Medrol) = m(Medrol) ÷ d(Medrol).
So, V(Medrol) = 49.39 mg ÷ 20 mg/mL = 2,469 mL.
2C6H14 + 13O2 ---> 6CO2 +14H2O
Calculating the molar mass of C6H14: M(C6H14)=12.011*6 +1.008*14 ≈ 86.17 g/mol
Thus, 86.17 g of C6H14 corresponds to 1 mole.
2C6H14 + 13O2 ---> 6CO2 +14H2O
based on the equation 2 mol 6 mol
according to the question 1 mol 3 mol
To determine M(CO2): M(CO2)= 12.011 + 2*15.999= 44.009 g/mol
Therefore, 3 mol CO2*44.009 g/1 mol CO2 ≈ 132.0 g CO2
Final answer: 132.0 g CO2
Answer:
The salt identified is barium chloride.
Explanation:
The moles of barium sulfate produced are 
per the reaction, 1 mole of barium sulfate arises from 1 mole of 
.
Therefore, 0.0480 moles result from:
of .
The quantity of used amounts to 10.00 g
Moles of = \frac{10.00 g}{\text{Molar mass}}[/tex]
The molar mass of is 208.33 g/mol
The closest answer to our calculation is 
.
The correct identification is barium chloride, which has a molar mass of 208.2 g/mol.
