Answer:
[H+] is 3.31 × 10^-11 M
Solution: Basic
Explanation:
In the question we are given;
Concentration of Hydroxide ion [OH-] as 3.0 × 10^-4 M
We are required to determine the [H+] and whether the solution is acidic, basic or neutral.
We know that;
pOH + pH = 14
and that;
pOH = - log [OH-]
Therefore;
pOH = - log ( 3.0 × 10^-4)
= 3.52
But;
pH = 14 - pOH
Thus;
pH = 14 - 3.52
= 10.48
But, we know that;
pH = - log[H+]
Therefore;
H+ = Antilog -pH
= Antilog -10.48
= 3.31 × 10^-11
Thus, the concentration of H+ ions [H+] is 3.31 × 10^-11 M
Also since the pH is 10.48, then the solution is basic
We are given the equation 3CO(g) + Fe2O3(s)= Fe(s) + 3CO2(g) and determine the next step in balancing the equation. In this case,C is already balanced. The next step is to balance Fe so we put 2 before Fe (s) to balance the 2 atoms in the left of the equation.
Answer:
46 g
Explanation:
The balanced equation of the reaction between O and NO is
2 NO + O₂ ⇔ 2 NO₂
Now, you need to find the limiting reagent. Find the moles of each reactant and divide the moles by the coefficient in the equation.
NO: (80 g)/(30.006 g/mol) = 2.666 mol
(2.666 mol)/2 = 1.333
O₂: (16 g)/(31.998 g/mol) = 0.500 mol
(0.500 mol)/1 = 0.500 mol
Since O₂ is smaller, this is the limiting reagent.
The amount of NO₂ produced will depend on the limiting reagent. You need to look at the equation to determine the ratio. For every mole of O₂ reacted, 2 moles of NO₂ are produced.
To find grams of NO₂ produced, multiply moles of O₂ by the ratio of NO₂ to O₂. Then, convert moles of NO₂ to find grams.
0.500 mol O₂ × (2 mol NO₂/1 mol O₂) = 1.000 mol NO₂
1.000 mol × 46.005 g/mol = 46.005 g
You will produce 46 g of NO₂.
Explanation:
Radium (atomic number 88) has similar properties to barium and is also in the Group 2 category. However, radium is a radioactive element and is generally under the category of radioisotopes in addition to being an alkaline earth metal, because it is not a stable element.
