What does it mean when two solid objects take up different spaces?

For a reaction A + B → products, the following data were collected. Experiment Number Initial Concentration of A (M) Initial Con

Afina-wow [57]

Answer:

Rate constant k = 1.57*10⁻⁵ s⁻¹

Explanation:

Given reaction:

$A\rightarrow B$

Expt [A] M [B] M Rate [M/s]

1 3.40 4.16 1.82*10^-4

2 4.59 4.16 3.32*10^-4

3. 3.40 5.46 1.82*10^-4

$Rate = k[A]^{x}[B]^{y}$

where k = rate constant

x and y are the orders wrt to A and B

To find x:

Divide rate of expt 2 by expt 1

$\frac{3.32*10^{-4} }{1.82*10^{-4} } =\frac{[4.59]^{x} [4.16]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\x =2$

To find y:

Divide rate of expt 3 by expt 1

$\frac{1.82*10^{-4} }{1.82*10^{-4} } =\frac{[3.40]^{x} [5.46]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\y =0$

Therefore: x = 2, y = 0

$Rate = k[A]^{2}[B]^{0}$

To find k

Use rate for expt 1:

$k = \frac{Rate1}{[A]^{2} } =\frac{1.82*10^{-4}M/s }{[3.40]^{2} } =1.57*10^{-5} s-1$

6 0

3 years ago

A) what is the light source for a plant growing in the shade?

Sholpan [36]

a.) Plants that thrive in the shade are often able to hold on to sunlight for extensive periods of time; they're in a sense like the camels of the plaNt WoRld.

b.) Though artificial lights are not nearly as beneficial as the sun, one could invest in one of them plant growing light thingies, but sun-loving plants might be sad if u do this instead of letting them soak in ePic rays from the sun.

4 0

3 years ago

A sucrose solution is prepared to a final concentration of 0.210 MM . Convert this value into terms of g/Lg/L, molality, and mas

vova2212 [387]

Answer:

1) 71.9 g/L

2) 0.221 m olal

3) 7.05% by mass

Explanation:

Step 1: Data given

Concentration of sucrose = 0.210 M

Molar weight of sucrose = 342.3 g/mol

Density of solution = 1.02 g/mL

Mass of water = 948.1 grams

Step 2: Convert this value into terms of g/L

(0.210 mol/L) * (342.3 g/mol) = 71.9 g/L

Calculate the molality

Step 1: Calculate mass water

Suppose we have a volume of 1.00L

Mass of the solution = 1000 mL * 1.02 g/mL = 1020 g solution

We know that there are 71.9 g of solute in a liter of solution from the first calculation. This means

(1020 grams solution) - (71.9 g solute) = 948.1 g = 0.9481 kg water

Step 2: Calculate molality

Molality = moles sucrose / mass water

(0.210 mol) / (0.9481 kg) = 0.221 mol/kg = 0.221 m olal

Mass %

% MAss = (mass solute / mass solution)*100%

(71.9 g) / (1020 g) *100% = 7.05% by mass

4 0

4 years ago

Which of the following statements is true about atoms?

Leokris [45]

Atoms of the same element may have different mass numbers.

6 0

3 years ago

Experiment #1A melting point of an old sample of Naphthalene was completed and a melting range of 77-83 oC was observed and reco

White raven [17]

Answer:

%error = 0.32%

Explanation:

Let's answer both questions, by parts.

1. Percentage error:

In this case, I do not have the video, but I do have the reported melting point of naphtalene which is 80.26 °C.

The expression to calculate the percentage error is the following:

%Error = absolute error / actual percentage. (1)

And the absolute error is:

Abs error = actual value - experimental value (2)

But the experimental value is a range, so we just have to get a average of that:

Exp value = 77 + 83 / 2 = 80 °C

Now the absolute error:

Abs error = 80.26 - 80 = 0.26 °C

Finally the %error:

%error = (0.26 / 80.26) * 100

<h2>%error = 0.32%</h2>

2. Meaning of melting point range and %error

The melting point range just means that the sample of naphtalene has impurities, and when a sample of any compound has impurities, melting point tends to be low. However, this decrease of temperature is a wider range. But usually a range of just 5° C means that compound has little traces of impurities but it can still be used for reactions.

The %error means that the impurities of the sample are really low, so the sample is practically pure with little traces of impurities.

3 0

3 years ago