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3 years ago

Why do people describe water as a universal solvent

Chemistry

2 answers:

dedylja [7]3 years ago

6 0

Water is capable of dissolving a variety of different substances

NikAS [45]3 years ago

5 0

The reason people describe water as a universal solvent is because it dissolves almost anything.

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Upper n subscript 2 (g) plus 3 upper H subscript 2 (g) double-headed arrow 2 upper N upper H subscript 3 (g). At equilibrium, th

artcher [175]

Answer:

The <u>equilibrium constant</u> is:

$k_c=0.0030M^{-2}$

Explanation:

The correct equation is:

N₂(g) + 3H₂(g) ⇄ 2NH₃(g)

Thus, with the equilibrium concentrations you can calculate the equilibrium constant, Kc.

The equation for the equilibrium constant is:

$k_c=\dfrac{[NH_3]^2}{[N_2]\cdot [H_2]^3}$

Substituting:

$k_c=\dfrac{(0.105M)^2}{(1.1M)\cdot (1.50M)^3}$

$k_c=0.0030M^{-2}$

6 0

3 years ago

Given the reaction: 4A1 + 302 - 2A1,O, How many grams of Al will be

dlinn [17]

Answer:

The answer to your question is 27 g of Al

Explanation:

Data

mass of Al = ?

moles of Al₂O₃ = 0.5

The correct formula for the product is Al₂O₃

Balanced chemical reaction

4Al + 3O₂ ⇒ 2Al₂O₃

Process

1.- Calculate the molar mass of the product

Al₂O₃ = (27 x 2) + (16 x 3)

= 54 + 48

= 102 g

2.- Convert the moles of Al₂O₃ to grams

102 g ---------------- 1 mol

x ---------------- 0.5 moles

x = (0.5 x 102) / 1

x = 51 g of Al₂O₃

3.- Use proportions to calculate the mass of Al

4(27) g of Al --------------- 2(102) g of Al₂O₃

x --------------- 51 g

x = (51 x 4(27)) / 2(102)

x = 5508 / 204

x = 27 g of Al

4 0

3 years ago

43. A stock glucose standard has a concentration of 1,000 mg/dL. A 1/5 dilution of this standard is made. What would be the fina

Nat2105 [25]

The final concentration of the diluted standard is 0.2 mg/dL.

<h3 /><h3>What is concentration of glucose standard after 1/5 solution?</h3>

Using the dilution formula:

C1V1 = C2V2

where

C1 is initial concentration
V1 initial volume
C2 is final concentration
V2 is final volume.

Assuming a final volume of 100 mL, and since a 1/5 dilution is made:

C1 = 1.00 mg/dL

V1 = 20

C2 = ?

V2 = 100 mL

C2 = C1V1/V2

C2 = 20 × 1/100

C2 = 0.2 mg/dL

Therefore, the final concentration of the diluted standard is 0.2 mg/dL.

Learn more about dilution at: brainly.com/question/24881505

6 0

2 years ago

Polar bonds share electrons _______.

Stells [14]

A polar bonds share electrons meaning they are a convenient bond

8 0

3 years ago

At a certain temperature the equilibrium constant, Kc, equals 0.11 for the reaction:

Butoxors [25]

<u>Answer:</u> The equilibrium concentration of ICl is 0.27 M

<u>Explanation:</u>

We are given:

Initial moles of iodine gas = 0.45 moles

Initial moles of chlorine gas = 0.45 moles

Volume of the flask = 2.0 L

The molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}$

Initial concentration of iodine gas = $\frac{0.45}{2}=0.225M$

Initial concentration of chlorine gas = $\frac{0.45}{2}=0.225M$

For the given chemical equation:

$2ICl(g)\rightarrow I_2(g)+Cl_2(g);K_c=0.11$

As, the initial moles of iodine and chlorine are given. So, the reaction will proceed backwards.

The chemical equation becomes:

$I_2(g)+Cl_2(g)\rightarrow 2ICl(g);K_c=\frac{1}{0.11}=9.091$

<u>Initial:</u> 0.225 0.225

<u>At eqllm:</u> 0.225-x 0.225-x 2x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[ICl]^2}{[Cl_2][I_2]}$

Putting values in above equation, we get:

$9.091=\frac{(2x)^2}{(0.225-x)\times (0.225-x)}\\\\x=0.135,0.668$

Neglecting the value of x = 0.668 because equilibrium concentration cannot be greater than the initial concentration

So, equilibrium concentration of ICl = 2x = (2 × 0.135) = 0.27 M

Hence, the equilibrium concentration of ICl is 0.27 M

6 0

3 years ago