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kolbaska11 [484]
2 years ago
14

What is the mass of 16.3 l of helium gas?

Chemistry
1 answer:
erma4kov [3.2K]2 years ago
5 0
 4.003 is the mass of helium gas


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Which of the following compounds would be most effective in lowering the melting point of ice on roads? a) CaCl2 b) NaCl C) K3PO
Mrac [35]
I'm not 100% sure on this, but I would go with C) NaCl. 
NaCl is a salt, and that is used to melt the ice on the roads. Hope this helps!
7 0
2 years ago
A 360mg sample of aspirin, C9H8O4, (molar mass 180g), is dissolved in enough water to produce 200mL of solution. What is the mol
notsponge [240]

Answer:

Molarity is 0.04M

Explanation:

First of all, let's determinate the moles of aspirin in that sample

Mass / Molar mass = Moles

360 mg = 0.360 g

0.360 g / 180 g/m = 0.002 moles

This moles that are included in 200 mL of solution, are also in 50 mL.

So molarity is mol/L

50 mL = 0.05 L

0.002 m / 0.05 L = 0.04M

6 0
2 years ago
In 1930 the american physicist ernest lawrence designed the first cyclotron in berkeley, california. in 1937 lawrence bombarded
gregori [183]
In, 1937 Lawrence, in operating his cyclotron, bombarded a molybdenum-96 foil with deuterium ions (2h), producing for the first time an element not found in nature. He was initially unaware that the radioactivity produced by the "bombarded foil" was not from molybdenum but from a new, artificial element. It was his cooperation with Italian-American physicist <span>Emilio Segrè </span>that allowed the new element to be discovered. The answer is Technetium: Tc
8 0
2 years ago
At 4.00 L, an expandable vessel contains 0.864 mol of oxygen gas. How many liters of oxygen gas must be added at constant temper
fgiga [73]

Answer:

We have to add 2.30 L of oxygen gas

Explanation:

Step 1: Data given

Initial volume = 4.00 L

Number of moles oxygen gas= 0.864 moles

Temperature = constant

Number of moles of oxygen gas increased to 1.36 moles

Step 2: Calculate new volume

V1/n1 = V2/n2

⇒V1 = the initial volume of the vessel = 4.00 L

⇒n1 = the initial number of moles oxygen gas = 0.864 moles

⇒V2 = the nex volume of the vessel

⇒n2 = the increased number of moles oxygen gas = 1.36 moles

4.00L / 0.864 moles = V2 / 1.36 moles

V2 = 6.30 L

The new volume is 6.30 L

Step 3: Calculate the amount of oxygen gas we have to add

6.30 - 4.00 = 2.30 L

We have to add 2.30 L of oxygen gas

4 0
2 years ago
One of the emission spectral lines for Be31 has a wavelength of 253.4 nm for an electronic transition that begins in the state w
max2010maxim [7]

Explanation:

The given data is as follows.

   \lambda = 253.4 nm = 253.4 \times 10^{-9}m      (as 1 nm = 10^{-9})

            n_{1} = 5,        n_{2} = ?

Relation between energy and wavelength is as follows.

                    E = \frac{hc}{\lambda}

                       = \frac{6.626 \times 10^{-34} Js \times 3 \times 10^{8} m/s}{253.4 \times 10^{-9}}

                       = 0.0784 \times 10^{-17} J

                       = 7.84 \times 10^{-19} J

Hence, energy released is 7.84 \times 10^{-19} J.

Also, we known that change in energy will be as follows.

     \Delta E = -2.178 \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}

where, Z = atomic number of the given element

 7.84 \times 10^{-19} J = -2.178 \times (4)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}

    \frac{7.84 \times 10^{-19} J}{34.848} = \frac{1}{n^{2}_{2}} - \frac{1}{(5)^{2}}

      0.02 + 0.04 = \frac{1}{n^{2}_{1}}

                      n_{1} = \sqrt{\frac{1}{0.06}}

                          = 4

Thus, we can conclude that the principal quantum number of the lower-energy state corresponding to this emission is n = 4.

4 0
2 years ago
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