I was born in ……… west.<br><br>complete this sentense with article

You want to prepare 500.0 mL of 1.000 M KNO3 at 20°C, but the lab (and water) temperature is 24°C at the time of preparation. Ho

timama [110]

Explanation:

As per the given data, at a higher temperature, at $24^{o}C$ , the solution will occupy a larger volume than at $20^{o}C$ .

Since, density is mass divided by volume and it will decrease at higher temperature.

Also, concentration is number of moles divided by volume and it decreases at higher temperature.

At $20^{o}C$ , density of water=0.9982071 g/ml

Therefore, $\frac{concentration}{density}$ will be calculated as follows.

= $\frac{C_{1}}{d_{1}}$

= $\frac{1.000 mol/L}{0.9982071 g/ml}$

= 1.0017961 mol/g

At $24^{o}C$ , density of water = 0.9972995 g/ml

Since, $\frac{concentration}{density}$ = $\frac{C_{2}}{d_{2}}$

= $\frac{C_{2}}{0.9972995}$

Also, $\frac{C_{1}}{d_{1}}$ = $\frac{C_{2}}{d_{2}}$

so, 1.0017961 mol/g = $\frac{C_{2}}{0.9972995}$

$C_{2} = 1.0017961 \times 0.9972995$

= 0.9990907 mol/L

Therefore, in 500 ml, concentration of $KNO_{3}$ present is calculated as follows.

$C_{2}$ = $\frac{concentration}{volume}$

0.9990907 mol/L = $\frac{concentration}{0.5 L}$

concentration = 0.49954537 mol

Hence, mass (m'') = $0.49954537 mol \times 101.1032 g/mol$ = 50.5056 g (as molar mass of $KNO_{3}$ = 101.1032 g/mol).

Any object displaces air, so the apparent mass is somewhat reduced, which requires buoyancy correction.

Hence, using Buoyancy correction as follows,

m = $m''' \times \frac{(1 - \frac{d_{air}}{d_{weights}})}{(1 - \frac{d_{air}}{d})}}$

where, $d_{air}$ = density of air = 0.0012 g/ml

$d_{weight}$ = density of callibration weights = 8.0g/ml

d = density of weighed object

Hence, the true mass will be calculated as follows.

True mass(m) = $50.5056 \times \frac{(1 - \frac{0.0012}{8.0})}{(1 - (\frac{0.0012}{2.109})}$

true mass(m) = 50.5268 g

= 50.53 g (approx)

Thus, we can conclude that 50.53 g apparent mass of $KNO_{3}$ needs to be measured.

8 0

3 years ago

Describe Newton's Law of Viscosity and the constitutive behavior of non-Newtonian fluid

CaHeK987 [17]

Answer:

- "Newton’s viscosity law’s states that, the shear stress between adjacent fluid layers is proportional to the velocity gradients between the two layers".

- A non-Newtonian fluid is a fluid which the relationship between the shear stress and the velocity gradient is not properly defined by the Newton's viscosity law, thus, the behavior is not lineal but potential.

Explanation:

Hello, here the answers:

- "Newton’s viscosity law’s states that, the shear stress between adjacent fluid layers is proportional to the velocity gradients between the two layers" (taken from Kundu, P. K., Cohen, I. M., & Dowling, D. R. (2012). Fluid mechanics.), thus, it means that when you have a fluid with an acting-on-it share stress (an external force which move the fluid), the related velocity gradient (variation or change in velocity) at which the layers are moving are related as:

$\pi =-v \frac{du}{dy}$

Whereas $\pi$ is the shear stress, $v$ is the viscosity and the differential accounts for the change in the velocity in the arbitrary $y$ coordinate.

- A non-Newtonian fluid is a fluid which the relationship between the shear stress and the velocity gradient is not properly defined by the Newton's viscosity law, thus, the behavior is not lineal but potential, based on:

$\pi =-v (\frac{du}{dy})^n$

Whereas $n$ accounts for a decreasing or increasing behavior of the shear stress.

Best regards.

8 0

3 years ago

How many moles of gas would occupy 22.4 L at 273K in one arm?

andrew-mc [135]

Answer:

1 mol

Explanation:

Using the general gas law equation as follows:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

According to the provided information in the question;

V = 22.4L

T = 273K

P = 1 atm

R = 0.0821 Latm/molK

n = ?

Using PV = nRT

n = PV/RT

n = (1 × 22.4) ÷ (0.0821 × 273)

n = 22.4 ÷ 22.4

n = 1mol

4 0

3 years ago

Which of the following 0.820 M solutions would have the greatest colligative effect?

eimsori [14]

Answer:

K3PO4

Explanation:

Recall that colligative properties depends on the number of particles present. The greater the number of particles present, the greater the degree of colligative properties of the solution. Let us look at each option individually;

SrCr2O7-------> Sr^2+ + Cr2O7^2- ( 2 particles)

C4H11N (not ionic in nature hence it can not dissociate into ions)

K3PO4-------> 3K^+ + PO4^3- (4 particles)

Rb2CO3-------> 2Rb^+ + CO3^2- (3 particles)

Hence K3PO4 has the greatest number of particles and will display the greatest colligative effect.

8 0

3 years ago

How many valence electrons does Rb (rubidium) have?

Murljashka [212]

Answer:

The total number of electrons in a valence shell is called valence electrons. The electron configuration of rubidium shows that the last shell of rubidium has an electron. Therefore, the valence electrons of rubidium(Rb) are one.

3 0

2 years ago

I was born in ……… west.complete this sentense with article​

I was born in ……… west.complete this sentense with article