Answer:
P₂ = 1.0 atm
Explanation:
Boyles Law problem => P ∝ 1/V at constant temperature (T).
Empirical equation
P ∝ 1/V => P = k(1/V) => k = P·V => for comparing two different case conditions, k₁ = k₂ => P₁V₁ = P₂V₂
Given
P₁ = 1.6 atm
V₁ = 312 ml
P₂ = ?
V₂ = 500 ml
P₁V₁ = P₂V₂ => P₂ = P₁V₁/V₂ =1.6 atm x 312 ml / 500ml = 1.0 atm
This compound is Boron selenate. Molar mass of B2(SeO4)3 is 450.4948 g/mol.
A 1.775g sample mixture of KHCO₃ is decomposed by heating. if the mass loss is 0.275g, the mass percentage of KHCO₃ is 70.4%.
<h3>What is a decomposition reaction?</h3>
A decomposition reaction can be defined as a chemical reaction in which one reactant breaks down into two or more products.
- Step 1: Write the balanced equation for the decomposition of KHCO₃.
2 KHCO₃(s) → K₂CO₃(s) + CO₂(g) + H₂O(l)
The mass loss of 0.275 g is due to the gaseous CO₂ that escapes the sample.
- Step 2: Calculate the mass of KHCO₃ that formed 0.275 g of CO₂.
In the balanced equation, the mass ratio of KHCO₃ to CO₂ is 200.24:44.01.
0.275 g CO₂ × 200.24 g KHCO₃/44.01 g CO₂ = 1.25 g KHCO₃
- Step 3: Calculate the mass percentage of KHCO₃ in the sample.
There are 1.25 g of KHCO₃ in the 1.775 g sample.
%KHCO₃ = 1.25 g/1.775 g × 100% = 70.4%
A 1.775g sample mixture of KHCO₃ is decomposed by heating. if the mass loss is 0.275g, the mass percentage of KHCO₃ is 70.4%.
Learn more about decomposition reactions here: brainly.com/question/14219426
Answer:
Reactive and lose 1 electron
Explanation:
We know that the element Z = 119 would be placed right below the Fr, in the column of the alcaline metals.
We also know that the trend in the electronegativity is to decrease when you go up-down ia group.
The known electronegativities of the elements of this group are:
Li: 0.98
Na: 0.93
K: 0.82
Rb: 0.82
Cs: 0.79
Fr: 0.70
Then the hypotetical element Z = 119 would probably have an electronegativity slightly below 0.70, for sure in the range 0.60 - 0.70.