Determine the molar mass for ammonia (NH3) in g/mol.

Exactly 5.00 L of air at -50.0°C is warmed to 100.0°C. What is the new volume if the pressure remains constant?

zaharov [31]

Explanation:

The given data is as follows.

$V_{1}$ = 5.00 L, $V_{2}$ = ?

$T_{1}$ = -50.0 + 273 = 223 K, $T_{2}$ = 100.0 + 273 = 373 K

P = constant

Therefore, calculate $V_{2}$ as follows.

$\frac{PV_{1}}{T_{1}}$ = $\frac{PV_{2}}{T_{2}}$

Since pressure is constant so, P will be cancelled out from both the sides.

$\frac{V_{1}}{T_{1}}$ = $\frac{V_{2}}{T_{2}}$

$V_{2}$ = $\frac{V_{1}T_{2}}{T_{1}}$

= $\frac{5.00 L \times 373 K}{223 K}$

= $\frac{1865 L}{223}$

= 8.36 L

Thus, we can conclude that $V_{2}$ is 8.36 L.

3 0

3 years ago

What is the lowest point of a wave called?<br><br> Amplitude<br> Crest<br> Frequency<br> Trough

zavuch27 [327]

<h3>Answer:</h3>

Trough

<h3>Explanation:</h3>

We need to answer the following questions

What is a wave?

A wave refers to a transmission of a disturbance from one point to another. It involves transfer of energy from the source to other points.

What are the major types of wave?

Waves may be classified as longitudinal waves or transverse wave based on the vibration of particles relative to the direction of transmission.

What is a transverse wave?

A transverse wave is a wave in which the vibration of particles is perpendicular to the direction of transmission of the wave.
This type of wave contains highest points known as crests and lowest points known as troughs.

5 0

3 years ago

Read 2 more answers

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,

Masja [62]

<u>Answer:</u> The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

<u>Explanation:</u>

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> $H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V$ ( × 3)

<u>Reduction half reaction:</u> $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

<u>Net reaction:</u> $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

Hence, the concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

7 0

3 years ago

If you had an object with a mass of 76g and a volume of 22 ml, what is its density? If you cut this object in half, what would b

V125BC [204]

D = M/V = 76g / 22ml = 3.4g/ml

Half ~ D = 38g / 11ml = 3.4g/ml

Even if the object you had was cut in half, it’s density would remain the same.

8 0

3 years ago

A group of students is investigating how the addition of salt impacts the floatation of an egg in water. Four identical cups wer

soldier1979 [14.2K]

Answer: The correct answer is Option 1.

Explanation:

Control group is the group used in the experiment by the researchers in which no change in the variable is done. It is then set as a benchmark for the groups which are being tested.

We are given 4 groups which are being experimented for the flotation of an egg in water. As in Cup 1, there is no addition of salt and hence there is no change in the variables. So, this is set as a benchmark fro the cups which are further used in the experiment conducted.

Hence, the correct answer is Option 1.

4 0

3 years ago