Lab activity - Law of Conservation of Mass with vinegar and baking soda.

The denser and thicker a liquid is, the higher the buoyancy of an object is. In other words, if the density of a certain fluid i

Marta_Voda [28]

Answer:

Yes this is true.

Explanation:

3 0

2 years ago

Aptitude is the capacity to learn a particular skill or acquire a particular body of knowledge. Please select the best answer fr

Luden [163]

<h2>Answer:</h2>

True

<h2>Explanation:</h2>

we can describe aptitude as an innate ability of a person to do something. There are different aptitude test in which we test the skills and ability to do different work or his attitude toward success or failure.

So, we can say that aptitude is the capacity to learn a particular skill or acquire a particular body of knowledge.

4 0

3 years ago

Read 2 more answers

An inverted pyramid is being filled with water at a constant rate of 45 cubic centimeters per second. The pyramid, at the top, h

vaieri [72.5K]

Answer:

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

Explanation:

Length of the base = l

Width of the base = w

Height of the pyramid = h

Volume of the pyramid = $V=\frac{1}{3}lwh$

We have:

Rate at which water is filled in cube = $\frac{dV}{dt}= 45 cm^3/s$

Square based pyramid:

l = 6 cm, w = 6 cm, h = 13 cm

Volume of the square based pyramid = V

$V=\frac{1}{3}\times l^2\times h$

$\frac{l}{h}=\frac{6}{13}$

$l=\frac{6h}{13}$

$V=\frac{1}{3}\times (\frac{6h}{13})^2\times h$

$V=\frac{12}{169}h^3$

Differentiating V with respect to dt:

$\frac{dV}{dt}=\frac{d(\frac{12}{169}h^3)}{dt}$

$\frac{dV}{dt}=3\times \frac{12}{169}h^2\times \frac{dh}{dt}$

$45 cm^3/s=3\times \frac{12}{169}h^2\times \frac{dh}{dt}$

$\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times h^2}$

Putting, h = 4 cm

$\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times (4 cm)^2}$

$=13.20 cm/s$

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

3 0

3 years ago

A) Find the gas speed of sulfur dioxide at 100.0 degrees Celsius? ______________

gtnhenbr [62]

a. 381.27 m/s

b. the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triiodide

<h3>Further explanation</h3>

Given

T = 100 + 273 = 373 K

Required

a. the gas speedi

b. The rate of effusion comparison

Solution

a.

Average velocities of gases can be expressed as root-mean-square averages. (V rms)

$\large {\boxed {\bold {v_ {rms} = \sqrt {\dfrac {3RT} {Mm}}}}$

R = gas constant, T = temperature, Mm = molar mass of the gas particles

From the question

R = 8,314 J / mol K

T = temperature

Mm = molar mass, kg / mol

Molar mass of Sulfur dioxide = 64 g/mol = 0.064 kg/mol

$\tt v=\sqrt{\dfrac{3\times 8.314\times 373}{0.064} }\\\\v=381.27~m/s$

b. the effusion rates of two gases = the square root of the inverse of their molar masses:

$\rm \dfrac{r_1}{r_2}=\sqrt{\dfrac{M_2}{M_1} }$

M₁ = molar mass sulfur dioxide = 64

M₂ = molar mass nitrogen triodide = 395

$\tt \dfrac{r_1}{r_2}=\sqrt{\dfrac{395}{64} }=\dfrac{20}{8}=2.5$

the rate of effusion of sulfur dioxide = 2.5 faster than nitrogen triodide

4 0

2 years ago

At 400 K, the rate of decomposition of a gaseous compound initially at a pressure of 12.6 kPa, was 9.71 Pa s-1 when 10.0 per cen

Neporo4naja [7]

Answer:

The order of the reaction with respect to the gas = 2

Explanation:

Let the original gas pressure be [G₀]

Initial rate of reaction is given as

r = k [G₀]ⁿ

When 10% had reacted, amount of gas left = [0.9G₀], r = 9.71 Pa/s

r = k [0.9G₀]ⁿ = 9.71 (eqn 1)

when 20% had reacted, amount of gas left = [0.8G₀], r = 7.67 Pa/s

r = k [0.8G₀]ⁿ = 7.67 (eqn 2)

Dividing (eqn 1) by (eqn 2)

(9.71/7.67) = [0.9/0.8]ⁿ

1.266 = 1.125ⁿ

1.125ⁿ = 1.266

Take natural logarithms of both sides

n (In 1.125) = In 1.266

n = 0.236/0.118

n = 2.

7 0

3 years ago