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3 years ago

Discuss the relationship between amount of energy and the color of light produced by excited electrons

Chemistry

1 answer:

djverab [1.8K]3 years ago

5 0

If you go to this site you should find your answer. http://www.brooklyn.cuny.edu/bc/ahp/LAD/C3/C3_elecEnergy.html

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An analytical chemist is titrating 181.2 mL of a 0.09000M solution of diethylamine ((CH) NH) with a 0.5400M solution of HNO3. Th

gavmur [86]

Answer:

874 por que mi pen3 ne tu ocncha

Explanation:

mkmkm

8 0

2 years ago

15 POINTS PLS HELP!!!!!

scZoUnD [109]

Answer: (C) Vaporizing

Explanation:

Vaporization is the process in which the substance change the state of of liquid into the gas state.

The vaporization process require the largest input of the energy as when the state is in the solid state then, the solid substances contain the strong forces of the attraction and they require high energy to break these strong bonds.

For changing the liquid state into the gases state we require to overcome the surface tension and require enough energy for acquiring the vaporization state.

Therefore, option (C) is correct.

5 0

3 years ago

Nicotine, a component of tobacco, is composed of C, H, and N. A 7.875-mg sample of nicotine was combusted, producing 21.363 mg o

Gnom [1K]

Answer: The empirical formula for the given compound is $C_5H_7N$

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen, and nitrogen follows:

$C_xH_yN_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and nitrogen respectively.

We are given:

Mass of $CO_2=21.363mg=21.363\times 10^3g=21363g$

Mass of $H_2O=6.125g=6.125\times 10^3g=6125g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 21363 g of carbon dioxide, $\frac{12}{44}\times 21363=5826.27g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 6125 g of water, $\frac{2}{18}\times 6125=680.55$ of hydrogen will be contained.

Now we have to calculate the mass of nitrogen.

Mass of nitrogen in the compound = (7875) - (5826.27 + 680.55) = 1368.18 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5826.27g}{12g/mole}=485.52moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{680.55g}{1g/mole}=680.55moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{1368.18g}{14g/mole}=97.73moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0154 moles.

For Carbon = $\frac{485.52}{97.73}=4.96\approx 5$

For Hydrogen = $\frac{680.55}{97.73}=6.96\approx 7$

For Nitrogen = $\frac{97.73}{97.73}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

Hence, the empirical formula for the given compound nicotine is $C_5H_7N_1=C_5H_7N$

7 0

2 years ago

n the reaction Mg (s) + 2HCl (aq) H2 (g) + MgCl2 (aq), how many moles of hydrogen gas will be produced from 75.0 milliliters of

Xelga [282]

0.075 L * 1.0 M = 0.075 mol HCl
H2 is half of HCl by the coefficients divide 0.075 by 2 and get 0.0375 mol H2 gas

8 0

3 years ago

Use the periodic table to match each of the following element symbols to its name, atomic mass, or atomic number

uysha [10]

Answer:

Explanation:

From the periodic table, the element designated as Sn is tin. Sn is derieved from a latin name of the metal called Stanum

Selenium Se is a group 6 element. It belongs to the same group with oxygen, sulfur and tellurium. This substance has an atomic weight of 78.96g/mol

Oxygen is a group 6 element with an atomic mass of 16. The atomic mass is the number of protons in the nucleus of this atom.

5 0

3 years ago

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