Answer:
18.0 Ampere is the size of electric current that must flow.
Explanation:
Moles of electron , n = 550 mmol = 0.550 mol
1 mmol = 0.001 mol
Number of electrons = N
Charge on N electrons : Q
Duration of time charge allowed to pass = T = 49.0 min = 49.0 × 60 seconds
1 min = 60 seconds
Size of current : I
18.0 Ampere is the size of electric current that must flow.
Answer:
Explanation:
Firstly, we have to determine the mass of metal X. We can do that by interpreting the first and second statement mathematically.
Metal X can form 2 oxides (A and B).
A + B = 3g
The mass of oxygen in A is 0.72g and the mass of oxygen in B is 1.16g.
The mass of metal X in the two oxides will be the same because it's the same metal.
Thus, we represent the mass of the metal in the two oxides as 2X.
2X + 0.72 + 1.16 = 3
2X + 1.88 = 3
2X = 3 - 1.88
2X = 1.12
X = 0.56
<u>Thus, 0.56 g of the metal combines with 0.72g of oxygen in A and 1.16 g of oxygen in B.</u>
Thus, mass of metal (X) in 1g of oxygen in A is
0.56g ⇒ 0.72g
X ⇒ 1
X = 1 × 0.56/0.72
X = 0.78 g
Hence, 0.78g of the metal will combine with 1g of oxygen for A
Also, mass of metal (X) in 1g of oxygen in B is
0.56g ⇒ 1.16g
X ⇒ 1g
X = 1×0.56/1.16
X = 0.48 g
Thus, 0.48g of the metal will combine with 1g of oxygen for B
Answer:
it is soluble in water
Explanation:
mark brainliest pliiz cutee:)
The molar concentration of the original HF solution : 0.342 M
Further explanation
Given
31.2 ml of 0.200 M NaOH
18.2 ml of HF
Required
The molar concentration of HF
Solution
Titration formula
M₁V₁n₁=M₂V₂n₂
n=acid/base valence (amount of H⁺/OH⁻, for NaOH and HF n =1)
Titrant = NaOH(1)
Titrate = HF(2)
Input the value :
Balance the equation first:
2 KClO3 (s) ---> 2 KCl (s) + 3 O2 (g)
Moles of KClO3 = 110 / 122.5 = 0.89
Following the balanced chemical equation:
We can say moles of O2 produce =
x moles of KClO3
So, O2 = (3 / 2) x 0.89
= 1.34 moles
So, Volume at STP = nRT / P
T = <span>273.15 K
P = 1 atm
So, V = (1.34 x 0.0821 x 273.15) / 1 = 30.2 L</span>
