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3 years ago

6

A student made a graph to show the chemical equilibrium position of a reaction. The student forgot to label the y-axis of the gr

aph.
What best explains the label that the student should use on the y-axis?

Reaction rate, as the rate of forward and back reaction become equal at equilibrium.

Reaction rate, as the rate of forward reaction increases and rate of backward reaction decreases with time.

Concentration, as the concentration of products decreases and that of reactants increase over time.

Concentration, as the concentrations of reactants and products remain constant after equilibrium is reached.

1 answer:

Kazeer [188]3 years ago

7 0

Answer:

D. concentration, as the concentrations of reactants and products remain unchanged after equilibrium is reached.

Explanation:

I put it for the test and i got right hehe

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Consider 100.0 g samples of two different compounds consisting only of carbon and oxygen. One compound contains 27.2 g of carbon

Pani-rosa [81]

<u>Answer:</u> The ratio of carbon in both the compounds is 1 : 2

<u>Explanation:</u>

Law of multiple proportions states that when two elements combine to form two or more compounds in more than one proportion. The mass of one element that combine with a given mass of the other element are present in the ratios of small whole number. For Example: $Cu_2O\text{ and }CuO$

<u>For Sample 1:</u>

Total mass of sample = 100 g

Mass of carbon = 27.2 g

Mass of oxygen = (100 - 27.7) = 72.8 g

To formulate the formula of the compound, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{27.2g}{12g/mole}=2.26moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{72.8g}{16g/mole}=4.55moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 2.26 moles.

For Carbon = $\frac{2.26}{2.26}=1$

For Oxygen = $\frac{4.55}{2.26}=2.01\approx 2$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 2

Hence, the formula for sample 1 is $CO_2$

<u>For Sample 2:</u>

Total mass of sample = 100 g

Mass of carbon = 42.9 g

Mass of oxygen = (100 - 42.9) = 57.1 g

To formulate the formula of the compound, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{42.9g}{12g/mole}=3.57moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{57.1g}{16g/mole}=3.57moles$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.57 moles.

For Carbon = $\frac{3.57}{3.57}=1$

For Oxygen = $\frac{3.57}{3.57}=1$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : O = 1 : 1

Hence, the formula for sample 1 is $CO$

In the given samples, we need to fix the ratio of oxygen atoms.

So, in sample one, the atom ratio of oxygen and carbon is 2 : 1.

Thus, for 1 atom of oxygen, the atoms of carbon required will be = $\frac{1}{2}\times 1=\frac{1}{2}$

Now, taking the ratio of carbon atoms in both the samples, we get:

$C_1:C_2=\frac{1}{2}:1=1:2$

Hence, the ratio of carbon in both the compounds is 1 : 2

8 0

3 years ago

Bicarbonate concentrate mixers may have a which are replaced on a routine basic.

Sidana [21]

Answer: True the bicarbonate mixture can help save time and few routine.

Explanation:

For the purpose of making dialysate for hemodialysis patient therapies a bicarbonate mixing and delivering systems designed to prepare a liquid sodium bicarbonate formulation comes in handy.

Certain systems like the SDS unit also allow for the transfer and distribution of acid concentrate solutions. We also provide stand-alone acid concentrate delivery systems using a variety of holding tanks and delivery methods.

A challenge for hemodialysis providers is to properly provide bicarbonate solution in a cost effective manner. Preparation and disinfection can be time-consuming and labor intensive.

Bicarbonate however can corrode certain metals and painted surfaces leaving your preparation area encrusted and grimy.

Furthermore, if not mixed properly, bicarbonate can negatively affect the dialysate solution.

The answer to the above is true the bicarbonate mixture can help save time and few routine.

3 0

3 years ago

In a 71.4 g sample of nahco 3 3 , how many grams of sodium are present?

sweet [91]

There are 19.5 g Na in 71.4 g NaHCO₃

Calculate the <em>molecular mass of NaHCO₃</em>.

1 Na = 1 × 22.99 u = 22.99 u

1 H = 1 × 1.008 u = 1.008 u

1 C = 1 × 12.01 u = 12.01 u

3 O = 3 × 16.00 u = <u>48.00 u </u>

TOTAL = 84.008 u

So, there are 22.99 g of Na in 84.008 g NaHCO₃.

∴ Mass of Na = 71.4 g NaHCO₃ × (22.99 g Na/84.008 g NaHCO₃) = 19.5 g Na

4 0

3 years ago

Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib

Dvinal [7]

<u>Answer:</u> The value of $K_c$ for the final reaction is $7.16\times 10^{25}$

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u> $H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1$

<u>Equation 2:</u> $HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2$

The net equation follows:

$S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c$

As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

$K_c=\frac{1}{K_1}\times \frac{1}{K_2}$

We are given:

$K_1=9.57\times 10^{-8}$

$K_2=1.46\times 10^{-19}$

Putting values in above equation, we get:

$K_c=\frac{1}{(9.57\times 10^{-8})}\times \frac{1}{(1.46\times 10^{-19})}=7.16\times 10^{25}$

Hence, the value of $K_c$ for the final reaction is $7.16\times 10^{25}$

5 0

3 years ago

A sample of O2 with an initial temperature of 50.0 oC and a volume of 105 L is cooled to -25 oC. The new pressure is 105.4 kPa a

Damm [24]

Answer:

71.92 kPa

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (kPa)

P2 = final pressure (kPa)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

T1 = 50°C = 50 + 273 = 323K

V1 = 105L

T2 = -25°C = -25 + 273 = 248K

P2 = 105.4 kPa

P1 = ?

V2 = 55.0 L

Using P1V1/T1 = P2V2/T2

P1 × 105/323 = 105.4 × 55/248

105P1/323 = 5797/248

0.325P1 = 23.375

P1 = 23.375 ÷ 0.325

P1 = 71.92 kPa

4 0

3 years ago