What does pressure measure

A 2.684-g sample of zinc oxide was reduced by hydrogen gas, resulting in 2. 156 g of pure zinc metal. Determine the empirical fo

Afina-wow [57]

The empirical formula of the initial zinc oxide is ZnO.

<h3>What is Empirical Formula?</h3>

The empirical formula of a compound represents the ratios of elements in a compound but not the actual numbers or arrangement of the atoms.

It is the lowest whole number ratio of the element in the compound.

<h3>How to find out the empirical formula?</h3>

Find out the given masses and molar masses of the elements

The molar mass of Zn = 65 gmol⁻¹

Given the mass of Zn = 2.156 g

The molar mass of Oxygen = 16 gmol⁻¹

The mass of Oxygen = Mass of a sample of zinc oxide - the mass of zinc metal

= (2.684 - 2.156) g

= 0.528 g

Find the number of moles of the elements in the compound

The number of moles is given by

$n = \frac{m}{M}$

where m = given mass and

M = Molar mass

Number of moles of Zinc = $\frac{2.156}{65}$ = 0.033 moles

Number of moles of Oxygen = $\frac{0.528}{16}$ = 0.033 moles

Find the simplest ratios of the elements in the compound. To find the ratios simply divide the number of moles by the lowest number of moles obtained.

Here, the number of moles is the same for both elements. Hence, the simplest ratio for Zn:O is 1:1.

Therefore, the empirical formula of zinc oxide is ZnO.

Learn more about the empirical formula:

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5 0

1 year ago

Balance the equation for ethane C3H6 burning in oxygen to form carbon dioxide and steam. _____C3H6 + ____ O2 ---> CO2 + ____H

aleksandrvk [35]

Answer:

2C3H6 + 9 O2 ---> 6 CO2 + 6 H2O

Explanation:

8 0

3 years ago

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide dissolved in molten cryolit

natka813 [3]

The given question incomplete, the complete question is:

In the Hall-Heroult process, a large electric current is passed through a solution of aluminum oxide (Al,03) dissolved in molten cryolite (Na, Alts).re in the reduction of the Al, o, to pure aluminum. Suppose a current of 1800. A is passed through a Hall-Heroult cell for 37.0 seconds. Calculate the mass of pure aluminum produced Be sure your answer has a unit symbol and the correct number of significant digits.

Answer:

The correct answer is 6.2114 grams.

Explanation:

Based on the given question, the value of current or I have given is 1800 amperes, the time given is 37 seconds, and there is a need to find the mass of the pure aluminum generated in the process. Mass or weight can be determined by using Faraday's first law equation, that is, w = MIt/nF.

Here, M is the atomic mass, w is the weight of the substance deposited, t is time, I is current, n is the number of moles of the electron, and F is the Faraday's constant, which is 96500 C. In the process mentioned in the question, aluminum oxide is reduced to give rise to pure aluminum, and in the process 3 electrons are gained. So, the value of n will be 3. The M or the atomic mass of Al is 27 gm per mole. Now putting the values in the equation we get,

w = 27*1800*37 / 3*96500

w = 1798200 / 289500

w = 6.2114 grams

Hence, pure aluminum produced in the process is 6.2114 grams.

7 0

3 years ago

Some radioactive nuclides have very short half-lives, for example, I-31 has a half-life of approximately 8 days. Pu-234, by comp

lorasvet [3.4K]

Answer:

Here's what I find.

Explanation:

Iodine-131

Iodine-131 is both a beta emitter and a gamma emitter.

$_{53}^{131}\text{I}\longrightarrow \, _{54}^{131}\text{Xe} +\, _{-1}^{0}\text{e} +\, _{0}^{0}\gamma$

About 90 % of the energy is β-radiation and 10 % is γ-radiation. Both forms are highly energetic.

The main danger is from ingestion. The iodine concentrates in thyroid gland, where the β-radiation destroys cells up to 2 mm from the tissues that absorbed it.

Both the β- and γ-radiation cause cell mutations that can later become cancerous. Small doses, such as those absorbed from the nuclear disasters in the Ukraine and Japan, can cause cancers years after the original iodine has disappeared.

Plutonium-239

Plutonium-239 is an alpha emitter.

$_{94}^{239}\text{U} \longrightarrow \, _{92}^{235}\text{Xe} + \, _{2}^{4}\text{He}$

Alpha particles cannot penetrate the skin, so external exposure isn't much of a health risk.

However, they are extremely dangerous when they are inhaled and get inside cells. They travel first to the blood or lymph system and later to the bone marrow and liver, where they cause up to 1000 times more chromosomal damage than beta or gamma rays.

It takes about 20 years for plutonium to be eliminated from the liver around 50 years for from the skeleton, so it has a long time to cause damage.

6 0

3 years ago

How many Ne atoms are contained in 32.0 g of the element?

Svetlanka [38]

Mass atomic of Ne=20.18 u
Therefore:
molar mass=20.18 g/1 mol

1 mole=6.022*10²³ particles (atoms or molecules)

Then: 6.022*10²³ atoms are contained in 20.18g

Now, We can solve this problem by the three rule.

6.022*10²³ atoms-------------------20.18 g
x------------------------------------------32 g

x=(6.022*10²³ atoms * 32 g)/20.18 g=9.55*10²³ atoms.

Answer: 9.55*10²³ Ne atoms are contained in 32 g of the element.

3 0

3 years ago