second compound
Let molar mass of x is = X
Let molar mass of y is = Y
Moles of x in second compound = Mass / molar mass = 7 / X
Moles of y in second compound = Mass / molar mass = 4.5 / Y
For second compound
7 / X : 4.5/ Y = 1:1
Therefore
X / Y = 7/4.5
Y / X = 4.5/ 7
The mass of x in first compound = 14g
moles of x in first compound = 14/X
Mass of y in first compound = 3
moles of y in first compound = 3 / Y
14 / X : 3/ Y = 14Y / 3X = 14 X 4.5 / 3 X 7 = 3 :1
Thus molar ratio in first compound = moles of x / Moles of y = 3:2
Formula = x3y
Answer:
The average of given values is 2.1221 ml
Explanation:
Given data:
Given measurements = 3.00 ml , 2.0 ml, 2.987 × 10⁻³ml , 3.4856 ml
Average value of given measurements = ?
Solution:
Formula:
Average value = sum of all measurement / total number of measurements
2.987 × 10⁻³ml = 0.002987 ml
Now we will put the values.
Average value = 3.00 ml + 2.0 ml + 0.002987 ml+ 3.4856 ml / 4
Average value = 8.488587 ml / 4
Average value = 2.1221 ml
The average of given values is 2.1221 ml
Since the total amount of valence electrons is 3, it is in group 13 in the periodic table..therefore, it is specified as Boron.✅
Au^2S^3+ 3H^2 = 2Au + 3H^2S
Answer:
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Explanation:
