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2 years ago

What is the difference between a transition metal and an f-element?

1 answer:

6 0

Answer: The transition elements are in the d-block, and in the d-orbital have valence electrons. They can form several states of oxidation and contain different ions. Inner transition elements are in the f-block, and in the f-orbital have valence electrons.

Explanation:

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What is the pH of a mixture of 0.042 M NaH2PO4 and 0.058 M Na2HPO4? Hint: The pKa of phosphate is 6.86.

AlekseyPX

Answer:

The pH value of the mixture will be 7.00

Explanation:

Mono and disodium hydrogen phosphate mixture act as a buffer to maintain pH value around 7. Henderson–Hasselbalch equation is used to determine the pH value of a buffer mixture, which is mathematically expressed as,

$pH=pK_{a} + log(\frac{[Base]}{[Acid]})$

According to the given conditions, the equation will become as follow

$pH=pK_{a} + log(\frac{[Na_{2}HPO_{4} ]}{[NaH_{2}PO_{4}]})$

The base and acid are assigned by observing the pKa values of both the compounds; smaller value means more acidic. NaH₂PO₄ has a pKa value of 6.86, while Na₂HPO₄ has a pKa value of 12.32 (not given, but it's a constant). Another more easy way is to the count the acidic hydrogen in the molecular formula; the compound with more acidic hydrogens will be assigned acidic and vice versa.

Placing all the given data we obtain,

$pH=6.86 + log(\frac{0.058}{0.042})$

$pH=7.00$

5 0

3 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

2 years ago

Use the exponential term in the Arrhenius equation to explain how temperature affects reaction rate.

WINSTONCH [101]

The Arrhenius equation describes the relation between the rate of reaction and temperature for many physical and chemical reactions

It is an expression that provides a relationship between the rate constant (of a chemical reaction), the absolute temperature.

The Arrhenius equation,

k = $zpe^{\frac{- Ea}{RT} }$ , where

k is the rate constant,

z is the collision factor,

p is the steric factor,

Ea is the activation energy,

R = 8.3245 $\frac{J}{mol. K}$ is the ideal gas constant, and,

T is the temperature.

The activation energy by definition, is the minimum energy (or threshold energy) required for two particles of reactants upon collision to form products.

The Arrhenius equation could also be written as:

⇒ k = $Ae\frac{-Ea}{RT}$ , where

⇒ A = zp, the Arrhenius factor.

Taking the neutral logarithm of both parties, we get:

⇒ In k = $\frac{-Ea}{RT}\frac{(1)}{(T)}$ + In A,

Assuming that, A is independent of temperature, when T is increased, the equilibrium constant k will also increase and therefore, the rate of the reaction will also increase.

To learn more about Arrhenius equation here

brainly.com/question/9786461

#SPJ4

6 0

2 years ago

Consider a mixture of air and gasoline vapor in a cylinder with a piston. The original volume is 40. cm3. If the combustion of t

maks197457 [2]

Answer:

The gases will expand until a volume of Vf= 10812 cm^3

Explanation:

Since the gas mixture is expanding at constant pressure

$W=\int\limits^{Vf}_{Vi} {P} \, dV =P(Vf-Vi)$