The SI unit of heat energy is British Thermal unit Joule all of the above Calorie

If there are 3.10 moles of O, how many moles of each of the compounds are present?

Burka [1]

Explanation:

The question pretty much requires us to find the amount of moles of each compounds based on the number of moles of O given.

H2SO4

1 mol of H2SO4 contains 4 mol of O

x mol of H2SO4 would contain 3.10 mol of O

x = 3.10 * 1 / 4 = 0.775 mol of H2SO4

C2H4O2

1 mol of C2H4O2 contains 2 mol of O

x mol of C2H4O2 would contain 3.10 mol of O

x = 3.10 * 1 / 2 = 1.55 mol of C2H4O2

NaOH

1 mol of NaOH contains 1 mol of O

x mol of NaOH would contain 3.10 mol of O

x = 3.10 * 1 / 1 = 3.10 mol of NaOH

3 0

3 years ago

What are the benefits and dangers of the Iron element in our daily life?

Digiron [165]

Benefits; helps our red blood cells transport oxygen all around our body

8 0

4 years ago

From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp

Nina [5.8K]

Answer: The concentration of $NH_3$ required will be 0.285 M.

Explanation:

To calculate the molarity of $NiC_2O_4$ , we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles of $NiC_2O_4$ = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M$

For the given chemical equations:

$NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}$

$Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9$

Net equation: $NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?$

To calculate the equilibrium constant, K for above equation, we get:

$K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48$

The expression for equilibrium constant of above equation is:

$K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}$

As, $NiC_2O_4$ is a solid, so its activity is taken as 1 and so for $C_2O_4^{2-}$

We are given:

$[[Ni(NH_3)_6]^{2+}]=0.016M$

Putting values in above equations, we get:

$0.48=\frac{0.016}{[NH_3]^6}}$

$[NH_3]=0.285M$

Hence, the concentration of $NH_3$ required will be 0.285 M.

7 0

3 years ago

If the products of a certain reaction are favored (i.e., the ΔG value of the reaction is negative), one would expect a significa

Over [174]

Answer:

positive H and negative S

Explanation:

For a reaction to be spontaneous, the absolute best combination is a negative Delta H and a positive Delta S. When they are both positive, the reaction is only spontaneous at higher temperatures. When they are both negative, the reaction is only spontaneous at lower temperatures. and again if a catalyst is added to the reaction, the activation energy is lowered because a lower-energy transition state is formed. The catalyst does not affect the energy of the reactants or products (and thus does not affect ΔG).

So from these discussions

Ea does not affect G value at all (whether +Ea or -Ea).

And for product to be formed the reaction should be spontaneous, where H is negative and S positive else the reaction will yield low product.

4 0

4 years ago

Read 2 more answers

How many moles in 2.21 x10E24 atoms of aluminum

lisov135 [29]

<h3>Answer:</h3>

3.67 mol Al

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

2.21 × 10²⁴ atoms Al

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

Set up: $\displaystyle 2.21 \cdot 10^{24} \ atoms \ Al(\frac{1 \ mol \ Al}{6.022 \cdot 10^{23} \ atoms \ Al})$
Divide: $\displaystyle 3.66988 \ mol \ Al$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

3.66988 mol Al ≈ 3.67 mol Al

7 0

3 years ago