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V125BC [204]
3 years ago
8

A 10 gram rubber ball is placed in a graduated cylinder filled with 15 mL of water. If the volume in the graduated cylinder incr

eases to 21 mL after the ball sinks to the bottom of the cylinder, what is the density of the rubber ball?
Chemistry
1 answer:
Zigmanuir [339]3 years ago
6 0
The density of the rubber ball is calculated by the mass of the rubber ball divided by the volume of the rubber ball. 
21mL-15mL=6mL which is the volume of the rubber ball 
D=10g/6mL 
D=1.666g/mL density of the rubber ball
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Fireworks that contain metallic salts such as sodium, strontium, and barium can generate bright colors. A technician investigate
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Answer:

See explanation below

Explanation:

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c = \lambda f

And the energy of each photon in electromangetic wave of specific frequency is

E=hf

Where h is planck's constant

We can see that the wavelength, which determines color and position of bright line in spectroscope corresponds directly to the energy of electromagnetic wave, and thus, the characteristic energy of atom.

4. From the answer above, you now know that electronic transition can emit electron with specific wavelength. In strontium, emission of colored light occurs by the relaxation of electrons in excited state such that the difference between excited energy level and ground level falls in the energy of visible light spectral range.

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8 0
3 years ago
. Dimethyl nitrosamine is a known carcinogen. It can be formed in the intestinal tract when digestive juices react with the nitr
8090 [49]

Answer: The empirical formula for the given compound is C_2H_6ON_2

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen 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=5.134g

Mass of H_2O=3.173g

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 5.134 g of carbon dioxide, \frac{12}{44}\times 5.134=1.4g of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 3.173 g of water, \frac{2}{18}\times 3.173=0.35g of hydrogen will be contained.

For calculating the mass of nitrogen:

As, 100 g of sample contains 37.87 % of nitrogen

So, 4.319 g of sample contains \frac{37.87}{100}\times 4.319=1.635g of nitrogen

For calculating the mass of oxygen:

Mass of oxygen in the compound = (4.319) - (1.4 + 0.35 + 1.635) = 0.934 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{1.4g}{12g/mole}=0.116moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.35g}{1g/mole}=0.35moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.934g}{16g/mole}=0.058moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{1.635g}{14g/mole}=0.116moles

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 0.058 moles.

For Carbon = \frac{0.116}{0.058}=2

For Hydrogen  = \frac{0.35}{0.058}=6.03\approx 6

For Oxygen  = \frac{0.058}{0.058}=1

For Oxygen  = \frac{0.116}{0.058}=2

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O : N = 2 : 6 : 1 :2

Hence, the empirical formula for the given compound is C_2H_6O_1N_2=C_2H_6ON_2

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Answer:

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Explanation:

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