Answer:
There is 5.56 g of gold for every 1 g of chlorine
Explanation:
The ratio is the relationship between two numbers, defined as the ratio of one number to the other. So, the ratio between two numbers a and b is the fraction 
You know that a compound has 15.39 g of gold for every 2.77 g of chlorine. This can be expressed by the ratio:
The proportion is the equal relationship that exists between two reasons and is represented by: 
This reads a is a b as c is a d.
To calculate the amount of gold per 1 g of chlorine, the following proportion is expressed:
Solving for the mass of gold gives:
mass of gold= 5.56 grams
So, <u><em>there is 5.56 g of gold for every 1 g of chlorine</em></u>
Answer:
Option b. 22 g of He will have the greatest volume at STP
Explanation:
In order to determine the volume, we apply the Ideal Gases Law equation:
P . V = n . R . T
V = n . R . T / P
R, T and P are the same in all the situation we must define n (number of moles).
The one that has the greatest number of moles will have the greatest volume at STP
22 g of Ne . 1mol / 20.1 g = 1.09 moles of Ne
22g of He . 1mol / 4 g = 5.5 moles of He
22 g of O₂ . 1mol / 32g = 0.68 moles of O₂
22 g of Cl₂ . 1mol / 70.9 g = 0.31 moles of Cl₂
Answer:
B) 7.7
Explanation:
For the reaction Ag2CO3(s) + CrO42‒(aq) → Ag2CrO4(s) + CO32‒(aq)
Kc = (CO₃²⁻) / (CrO₄²⁻)
and the Ksp given are
Ag₂CO₃ ⇒ 2 Ag⁺(aq) + CO₃²⁻(aq) Ksp₁ = (Ag⁺)²(CO₃²⁻)
Ag₂CrO₄ ⇒ 2 Ag⁺(aq)+ CrO₄²⁻(aq) Ksp₂ = (Ag⁺)²(CrO₄²⁻)
Where (...) indicate concentrations M
Notice if we divide the expressions for Ksp we get:
Ksp₁/Ksp₂ = (CO₃²⁻) / (CrO₄²⁻) = 8.5 x 10⁻¹² / 1.1 x 10⁻¹² = 7.7
which is the desired answer.
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