All categories

2 years ago

A sample of nitrogen monoxide occupies a volume of 400 mL at 100 °C and 720 mmHg.

1 answer:

6 0

The volume of nitrogen monoxide that occupy at STP is= 277 Ml

calculation
The volume is obtained using the combined gas law that is P1V1/T1= P2V2?
Where P1 = 720 MmHg
V1 = 400ml
T1= 100 +273= 373 K
At STP temperature = 273 K and pressure= 760 mm hg
therefore T2= 273 k
p2 = 760 mmhg
V2=?
make V2 the subject of the formula

V2= (T2 ×P1 ×V1)/(P2×T1)

V2 is therefore = (273k x720 mmhg x 400 ml)/(760 mmhg x373K) = 277 Ml

You might be interested in

Manganese, Mn, forms two ions, one with a 2+ charge and one with a 3+ charge. What is the formula for manganese (II) sulfide?

maria [59]

It would be MnSO4

The (II) lets you know it’s the form with a 2+ charge and Sulfate has a 2- charge

These will cancel out making it plain MnSO4

If it was manganese (iii) sulfide the answer would be Mn2(SO4)3

4 0

3 years ago

Describe the contributions made by the following scientists to the development of the periodic table.

Basile [38]

Answer:

Explanation:

dimitri Mendeleev......

6 0

2 years ago

Use the reaction below to balance the nuclear reaction equation.

Temka [501]

Answer:

neon-22

Explanation:

i got it right..have a great day.Jesus loves you<3

4 0

2 years ago

Read 2 more answers

2. Sitting on a bench top are several samples: lithium metal (d = 0.53 g/mL), gold (d = 19.3 g/mL), aluminum (d = 2.70 g/mL), an

Snezhnost [94]

Answer:

The sample of lithium occupies the largest volume.

Explanation:

Given the densities for the four elements, we have the expression $d=\frac{m}{V}$ that shows the relationship between mass and Volume to express the density of an element.

For each element we have:

$d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL$

$d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL$

$d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL$

$d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL$

The problem says that all the samples have the same mass, so:

$m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m$

it means that m is a constant

Now, solving for the Volume in each element and with m as a constant, we have:

$V_{lithium}=\frac{m}{d_{lithium}}$

$V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m$

$V_{lithium}=1.88\frac{mL}{g}*m$

$V_{gold}=\frac{m}{d_{gold}}$

$V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m$

$V_{gold}=5.18*10^{-2}\frac{mL}{g}*m$

$V_{aluminum}=\frac{m}{d_{aluminum}}$

$V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m$

$V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m$

$V_{lead}=\frac{m}{d_{lead}}$

$V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m$

$V_{lead}=8.85*10^{-2}\frac{mL}{g}*m$

If we assume m = 1g, we find that:

$V_{lithium}=1.88mL$

$V_{gold}=5.18*10^{-2}mL$

$V_{aluminum}=3.70*10^{-1}mL$

$V_{lead}=8.85*10^{-2}mL$

So we can see that the sample of lithium occupies the largest volume with 1.88mL

Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.

4 0

2 years ago

Which electron configuration represents the

Pachacha [2.7K]

This problem is asking for the electron configuration of an excited atom of gallium, which according to the group it is, its excited state will theoretically be Ga³⁺, since it is a metal and loses three or less electrons, that is why it turns out positive. After writing the electron configuration, we realize the answer must be 2-8-17-4 according to the following:

<h3>Electron configurations:</h3><h3 />

In chemistry, electron configurations allow us arrange the electrons of an element according to specific energy levels and orbitals. In such a case, it turns out possible to write Gallium's electron configuration in its ground-state as follows:

$1s^22s^22p^63s^23p^64s^23d^{10}4p^1$

However, since the given choices do not match with a charge of 3+ as formerly explained, we assume this gallium atom will be excited to 1+, due to the fact that the third energy level comprising 18 electrons, will give one to the fourth energy level, turning out in an electron configuration of:

$1s^22s^22p^63s^23p^64s^23d^{9}4p^2$

Which matches with 2-8-17-4 as two electrons are present in the first energy level, eight in the second one, now seventeen in the third one and four in the fourth one.

Learn more about electron configurations: brainly.com/question/5524513

5 0

2 years ago