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3 years ago

What is the molarity of 2 mol of calcium carbonate, CaCl2, dissolved in 1 L of solution?

Physics

1 answer:

Agata [3.3K]3 years ago

6 0

Answer:

2mol/L

Explanation:

Given parameters:

Number of moles of calcium carbonate = 2mol

Volume of solution = 1L

Unknown:

Molarity = ?

Solution:

The molarity of a solution is the number of moles of solute dissolved or contained in a solution:

Molarity = $\frac{number of moles }{volume }$

Molarity = $\frac{2}{1}$ = 2mol/L

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The diagram shows waves of sound travelling through the air. By which factor is the sound intensity decreased at 3 meters?

SIZIF [17.4K]

The diagram is missing; however, we know that the intensity of a sound wave is inversely proportional to the square of the distance from the source:
$I(r)= \frac{1}{r^2}$
where I is the intensity and r is the distance from the source.

We can assume for instance that the initial distance from the source is r=1 m, so that we put
$I= \frac{1}{r^2}= \frac{1}{(1)^2}=1$
The intensity at r=3 m will be
$I= \frac{1}{r^2}= \frac{1}{(3)^2}= \frac{1}{9}$
Therefore, the sound intensity has decreased by a factor $1/9$ .

8 0

3 years ago

Read 2 more answers

During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound. The time between seeing

hodyreva [135]

Answer:

d = 1700 meters

Explanation:

During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound. The time between seeing the lighting and hearing the sound was 5 second, t = 5 seconds

Speed of sound, v = 340 m/s (say)

Let d is the distance of the colliding cloud from the observer. The distance covered by the object. It is given by :

$d=v\times t$

$d=340\ m/s\times 5\ s$

d = 1700 meters

So, the distance of the colliding cloud from the observer is 1700 meters. Hence, this is the required solution.

8 0

3 years ago

A series AC circuit contains a resistor, an inductor of 150 mH, a capacitor of 5.00 mF, and a generator with DVmax 5 240 V opera

yanalaym [24]

Given:

Inductance, L = 150 mH

Capacitance, C = 5.00 mF

$\Delta V_{max}$ = 240 V

frequency, f = 50Hz

$I_{max}$ = 100 mA

Solution:

To calculate the parameters of the given circuit series RLC circuit:

angular frequency, $\omega$ = $2\pi f = 2\pi \times50 = 100\pi$

a). Inductive reactance, $X_{L}$ is given by:

$\X_{L} = \omega L = 100\pi \times 150\times 10^{-3} = 47.12\Omega$

$X_{L} = 47.12\Omega$

b). The capacitive reactance, $X_{C}$ is given by:

$\X_{C} = \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 50\times 5.00\times 10^{-3}} = 0.636\Omega$

$X_{C} = 0.636\Omega$

c). Impedance, Z = $\frac{\Delta V_{max}}{I_{max}} = \frac{240}{100\times 10^{-3}} = 2400\Omega$

$Z = 2400\Omega$

d). Resistance, R is given by:

$Z = \sqrt {R^{2} + (X_{L} - X_{C})}$

$2400^{2} = R^{2} + (47.12 - 0.636)^{2}$

$R = \sqrt {5757839.238}$

$R = 2399.5\Omega$

e). Phase angle between current and the generator voltage is given by:

$tan\phi = \frac{X_{L} - X_{C}}{R}$

$\phi =tan^{-1}( \frac{X_{L} - X_{C}}{R})$

$\phi =tan^{-1}( \frac{47.12 - 0.636}{2399.5}) = tan^{-1}{0.0.01937}$

$\phi = 1.11^{\circ}$

5 0

4 years ago

What pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c?

Ronch [10]

The pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.

Therefore, option A is correct option.

Given,

Mass m = 14g

Volume= 3.5L

Temperature T= 75+273 = 348 K

Molar mass of CO = 28g/mol

Universal gas constant R= 0.082057L

Number of moles in 14 g of CO is

n= mass/ molar mass

= 14/28

= 0.5 mol

As we know that

PV= nRT

P × 3.5 = 0.5 × 0.082057 × 348

P × 3.5 = 14.277

P = 14.277/3.5

P = 4.0794 atm

P = 4.1 atm.

Thus we concluded that the pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.

learn more about pressure:

brainly.com/question/22613963

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5 0

1 year ago

When must a psychological researcher debrief human test subjects?

myrzilka [38]

<span>Psychological researchers must debrief human test subjects </span><span>at the end of every experiment.

The current code of ethics in p</span>sychological research states that researchers absolutely must debrief human test subjects at the end of every study regardless or whether or not harm or deception was involved.

Debriefing a subject after a study is an essential opportunity for the researcher to explain the purpose and aim of the study to the subject, make sure the subject is not harmed or mentally disturbed, clarify why deception was used (if deception was involved) and overall, to clarify any questions or doubts the subject might have.

3 0

3 years ago