Answer : The limiting reactant is
Explanation : Given,
Mass of
= 30.0 g
Mass of
= 75.0 g
Molar mass of
= 44 g/mole
Molar mass of
= 32 g/mole
First we have to calculate the moles of
and
.


Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,

From the balanced reaction we conclude that
As, 5 mole of
react with 1 mole of 
So, 2.34 moles of
react with
moles of 
From this we conclude that,
is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Therefore, the limiting reactant is
D. 2- youre adding two negative charges to a neutral element
Answer: ![K_c=\frac{[CH_3Cl]\times [OH^-]}{[CH_3OH]\times [Cl^-]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3Cl%5D%5Ctimes%20%5BOH%5E-%5D%7D%7B%5BCH_3OH%5D%5Ctimes%20%5BCl%5E-%5D%7D)
Explanation:
Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients. Pure solids are assumed to have a concentration of 1.
The given balanced equilibrium reaction is:

The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CH_3Cl]\times [OH^-]}{[CH_3OH]\times [Cl^-]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3Cl%5D%5Ctimes%20%5BOH%5E-%5D%7D%7B%5BCH_3OH%5D%5Ctimes%20%5BCl%5E-%5D%7D)
Thus the equilibrium constant expression for this reaction is ![K_c=\frac{[CH_3Cl]\times [OH^-]}{[CH_3OH]\times [Cl^-]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3Cl%5D%5Ctimes%20%5BOH%5E-%5D%7D%7B%5BCH_3OH%5D%5Ctimes%20%5BCl%5E-%5D%7D)
Answer:
The answer is A.allele
Explanation:
An allele is a variant form of a gene. Some genes have a variety of different forms, which are located at the same position, or genetic locus, on a chromosome. Humans are called diploid organisms because they have two alleles at each genetic locus, with one allele inherited from each parent.
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Answer:
7.61 liters
Explanation:
From the question,
Applying Charles law formula,
V₁/T₁ = V₂/T₂...................... Equation 1
Where P₁ = initial volume of of air, V₂ = Final volume of air, T₁ = Initial Temperature of air, T₂ = Final Temperature of air.
make V₂ the subject of the equation
V₂ = (V₁×T₂)/T₁............... Equation 2
Given: V₁ = 5 liters, T₁ -35 °C = (-35+273) K = 238 K, T₂ = 89°C = (89+273) K = 362 K
Substitute these values into equation 2
V₂ = (5×362)238
V₂ = 1810/238
V₂ = 7.61 liters