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expeople1 [14]
3 years ago
7

What is the first element on the periodic table?

Chemistry
2 answers:
tresset_1 [31]3 years ago
7 0
Hydrogen.
I hope this helps.
Galina-37 [17]3 years ago
4 0


Hydrogen, H.

I hope this helps!

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Calculate the volume of a container that is 8.9 centimeters long, 5.2 centimeters wide, and 2.55 cm tall. Choose an appropriate
choli [55]
The answer is 118.014 cm cube
4 0
3 years ago
A 2.241-g sample of nickel reacts with oxygen to form 2.852 g of the metal oxide.
nlexa [21]

Answer:

The empirical formula is = NiO

Explanation:

Given that:- Mass of nickel = 2.241 g

Mass of the oxide formed = 2.852 g

Mass of the oxygen reacted = Mass of the oxide formed - Mass of nickel  = 2.852 g - 2.241 g = 0.611 g

Molar mass of nickel  = 58.6934 g/mol

Moles of nickel = \frac{2.241}{58.6934}\ mol = 0.03818 mol

Molar mass of oxygen  = 15.999 g/mol

Moles of nickel = \frac{0.611}{15.999}\ mol = 0.03818 mol

Taking the simplest ratio for Ni and O as:

0.03818 : 0.03818 = 1 : 1

<u>The empirical formula is = NiO </u>

4 0
3 years ago
Argon, which comprises almost 1% of the atmosphere, is approximately 27 times more abundant than CO 2, but does not contribute t
kotegsom [21]

Answer:

The gas argon does not reach a state of vibrational excitation when infrared radiation strikes this gas.

Explanation:

The dry atmosphere is composed almost entirely of nitrogen (in a volumetric mixing ratio of 78.1%) and oxygen (20.9%), plus a series of oligogases such as argon (0.93%), helium and gases of greenhouse effect such as carbon dioxide (0.035%) and ozone. In addition, the atmosphere contains water vapor in very variable amounts (about 1%) and aerosols.

Greenhouse gases or greenhouse gases are the gaseous components of the atmosphere, both natural and anthropogenic, that absorb and emit radiation at certain wavelengths of the infrared radiation spectrum emitted by the Earth's surface, the atmosphere and clouds . In the Earth's atmosphere, the main greenhouse gases (GHG) are water vapor (H2O), carbon dioxide (CO2), nitrous oxide (N2O), methane (CH4) and ozone (O3 ). There is also in the atmosphere a series of greenhouse gases (GHG) created entirely by humans, such as halocarbons (compounds containing chlorine, bromine or fluorine and carbon, these compounds can act as potent greenhouse gases in the atmosphere and they are also one of the causes of the depletion of the ozone layer in the atmosphere) regulated by the Montreal Protocol. In addition to CO2, N2O and CH4, the Kyoto Protocol sets standards regarding sulfur hexafluoride (SF6), hydrofluorocarbons (HFCs) and perfluorocarbons (PFCs).

The difference between argon and greenhouse gases such as CO2 is that the individual atoms in the argon do not have free bonds and therefore do not vibrate. As a consequence, it does not reach a state of vibrational excitation when infrared radiation strikes this gas.

6 0
3 years ago
A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
natali 33 [55]

Answer:

There are 0.93 g of glucose in 100 mL of the final solution

Explanation:

In the first solution, the concentration of glucose (in g/L) is:

15.5 g / 0.100 L = 155 g/L

Then a 30.0 mL sample of this solution was taken and diluted to 0.500 L.

  • 30.0 mL equals 0.030 L (Because 30.0 mL ÷ 1000 = 0.030 L)

The concentration of the second solution is:

155 \frac{g}{L} *\frac{0.030L}{0.500L}=9.3\frac{g}{L}

So in 1 L of the second solution there are 9.3 g of glucose, in 100 mL (or 0.1 L) there would be:

1 L --------- 9.3 g

0.1 L--------- Xg

Xg = 9.3 g * 0.1 L / 1 L = 0.93 g

8 0
3 years ago
The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold.
azamat

Answer:

(a) m_{gold}=7.322g

(b)

V_{gold}=0.379cm^3

V_{copper}=0.122cm^3

(c) \rho _{coin}=15.94g/cm^3

Explanation:

Hello,

(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign  as shown below:

m_{gold}=\frac{m_{tota}*karats}{24}=\frac{7.988g*22}{24}=7.322g

(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:

V_{gold}=\frac{m_{gold}}{\rho_{gold}} =\frac{7.322g}{19.32g/cm^3}=0.379cm^3

m_{copper}=7.988g-7.322g=1.09g\\V_{copper}=\frac{m_{copper}}{\rho_{copper}}=\frac{1.09g}{8.94g/cm^3}  \\\\V_{copper}=0.122cm^3

(c) Finally, the volume is computed by dividing the total mass over the total volume containing both gold and copper:

\rho _{coin}=\frac{m_{total}}{V_{gold}+V_{copper}}=\frac{7.988 g}{0.379cm^3+0.122cm^3}\\  \\\rho _{coin}=15.94g/cm^3

Best regards.

3 0
3 years ago
Read 2 more answers
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