The atomic mass of X₁ is 14 and the name of the element is Nitrogen.
The atomic mass of X₂ is 16 and the name of the element is Oxygen.
The atomic mass of X₃ is 32 and the name of the element is Sulfur.
The given parameters include:
weight C weight X
Compound 1 4.2 4.9
Compound 2 3.5 4.67
Compound 3 1.3 3.47
The atomic mass of the first compound (X₁) is calculated as follow;
let the total mass of X₁ and C = T
X₁ + C = T
X₁ + C = T
X₁ + 12 = 26
X₁ = 26 - 12
X₁ = 14
The atomic mass of X₁ is 14 and the name of the element is Nitrogen.
The atomic mass of the second compound (X₂) is calculated as follow;
let the total mass = P
The atomic mass of X₂ is 16 and the name of the element is Oxygen.
The atomic mass of the third compound (X₃) is calculated as follow;
Let the total mass = y
The atomic mass of X₃ is 32 and the name of the element is Sulfur.
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Answer: yellow
When you mix solutions of lead (II) nitrate and potassium iodide. The precipitate is yellow in colour and the compound is lead (II) Iodide. Pb (NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq). Yes, it is a double displacement reaction.
Explanation: The lead nitrate solution contains particles (ions) of lead, and the potassium iodide solution contains particles of iodide. When the solutions mix, the lead particles and iodide particles combine and create two new compounds, a yellow solid called lead iodide and a white solid called potassium nitrate. When a solution of potassium iodide is added to a solution of lead nitrate taken in a test tube, the precipitation of a yellowish solid is observed. This yellowish solid is lead iodide. Potassium nitrate is formed along with lead iodide. When lead nitrate is mix with potassium iodine, a precipitation reaction occurs and yellow precipitate of lead iodide is formed.
Answer: 0.1 decrease in 15 minutes
unit rate is 0.1 / 15 = 0.00667
Temperature is decreasing by 0.00667 degrees Celsius per minute
Explanation:
Hope this helps !
Answer:
pH = 8.477
Explanation:
∴ ni 0.3 - -
nf - 0.3 0.3
∴ ni 0.50 - 0.3
nf 0.5 - X X 0.3 + X
∴ Ka = 2.0 E-9 = ([H3O+]*[OBr-]) / [HOBr]
⇒ Ka = 2.0 E-9 = ((X/1L)*(0.3 + X)/1 L) / ((0.5 - X)/1L)
⇒ 2.0 E-9 = ( 0.3X + X² ) / (0.5 - X)
⇒ X² + 0.3X - 1 E-9 = 0
⇒ X = 4.333 E-9 M
according Henderson-Hauselbach:
- pH = pk + Log [A-] / [HA]
∴ [OBr-] = 0.3 mol/ 1 L + 4.333 E-9 M = 0.300 M
∴ [HOBr] = 0.5 mol / 1 L - 4.33 E-9 M = 0.500 M
∴ pKa = - log Ka = - Log ( 2.0 E-9 ) = 8.6989
⇒ pH = 8.6989 + Log ( 0.300/0.500 )
⇒ pH = 8.477