Cell division. Hope this helps :)
Answer:
Mass = 112.54 g
Explanation:
Given data:
Mass of copper = 18 g
How much copper(II) nitrate formed = ?
Solution:
Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag
Number of moles of copper:
Number of moles = mass/ molar mass
Number of moles = 18 g/ 29 g/mol
Number of moles = 0.6 mol
Now we will compare the moles of Cu with Cu(NO₃)₂ .
Cu : Cu(NO₃)₂
1 : 1
0.6 : 0.6
Mass of Cu(NO₃)₂ :
Mass = number of moles × molar mass
Mass = 0.6 mol × 187.56 g/mol
Mass = 112.54 g
Answer:
C) to show that atoms are conserved in chemical reactions
Explanation:
When writing a chemical reaction, we should always consider the Mass Conservation Law, which basically states that; in an isolated system; the total mass should remain constant, this is, the total mass of the reactives should be equal to the total mass of the products
For this case, we should add the apporpiate coefficients in order to be in compliance with this law:
2H₂ + O₂ → 2H₂O
So, we can check the above statement:
For reactives (left side):
4H
2O
For product (right side):
4H
2O
Answer:
I think this is it
Water can have two different densities if it has substances dissolved in it. ... When liquid water freezes it becomes solid water or ice, which is less dense than liquid water. The fact that solid water (ice) is less dense than liquid water is evident in the way ice floats in a glass of water.
Explanation:
Answer:
[Ag⁺] = 5.0x10⁻¹⁴M
Explanation:
The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:
Ksp(PbI₂) = [Pb²⁺] [I⁻]² = 1.4x10⁻⁸
Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷
The PbI₂ <em>just begin to precipitate when the product [Pb²⁺] [I⁻]² = 1.4x10⁻⁸</em>
<em />
As the initial [Pb²⁺] = 0.0050M:
[Pb²⁺] [I⁻]² = 1.4x10⁻⁸
[0.0050] [I⁻]² = 1.4x10⁻⁸
[I⁻]² = 1.4x10⁻⁸ / 0.0050
[I⁻]² = 2.8x10⁻⁶
<h3>[I⁻] = 1.67x10⁻³</h3><h3 />
So, as the [I⁻] concentration is also in the expression of Ksp of AgI and you know concentration in solution of I⁻ = 1.67x10⁻³M:
[Ag⁺] [I⁻] = 8.3x10⁻¹⁷
[Ag⁺] [1.67x10⁻³] = 8.3x10⁻¹⁷
<h3>[Ag⁺] = 5.0x10⁻¹⁴M</h3>
