CrO and Cr₂O₃ make up the simplest chromium oxide formula.
What name does Cr₂O₃ use?
- Chromium oxide (Cr₂O₃)sometimes referred to as chromium sesquioxide or chromic oxide, is a compound in which chromium is oxidized to a +3 state. Sodium dichromate is calcined with either carbon or sulfur to produce it.
- Eskolaite, a mineral that bears the name of the Finnish geologist Pentti Eskola, is a kind of chromium oxide green that may be found in nature. The metallic glassy green surface of this unusual material has an unsettling moss-like look that may be used to conceal oneself in the environment.
- Studies on humans have conclusively shown that chromium (VI) breathed is a potential carcinogen, increasing the likelihood of developing lung cancer. According to animal studies, chromium (VI) exposure by inhalation can result in lung cancers.
Learn more about chromium here:
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nuclear power--used to turn turbines...
fossil fuels--burned to provide energy that is....
renewable energy--energy that with come back after use
outlet--a device....
steam--nuclear reactors....
I'm not sure but I tried lol,lemme know if I'm wrong :D
The reaction between oxygen, O2, and hydrogen, H2, to produce water can be expressed as,
2H2 + O2 --> 2H2O
The masses of each of the reactants are calculated below.
2H2 = 4(1.01 g) = 4.04 g
O2 = 2(16 g) = 32 g
Given 1.22 grams of oxygen, we determine the mass of hydrogen needed.
(1.22 g O2)(4.04 g H2 / 32 g O2) = 0.154 g of O2
Since there are 1.05 grams of O2 then, the limiting reactant is 1.22 grams of oxygen.
<em>Answer: 1.22 g of oxygen</em>
Two steps by inspection 1 qt = 0.25 gallons, 13.6 g/mL = 13.6 kg/L
0.25 gallon x 3.785411784 L/gallon x 13.6 kg/L x 1 lb/0.45359237 kg = 28.374 lb.
Hope this Helps!
Answer:
The new force will be \frac{1}{100} of the original force.
Explanation:
In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.
That said, let's say that our equation for the initial force is:
![F = G\frac{m_1m_2}{R^2}The problem states that the distance decrease to 1/10 of the original distance, this means:[tex]R_2 = \frac{1}{10}R](https://tex.z-dn.net/?f=F%20%3D%20G%5Cfrac%7Bm_1m_2%7D%7BR%5E2%7D%3C%2Fp%3E%3Cp%3EThe%20problem%20states%20%20that%20%20the%20distance%20decrease%20to%201%2F10%20of%20the%20original%20distance%2C%20this%20means%3A%3C%2Fp%3E%3Cp%3E%5Btex%5DR_2%20%3D%20%5Cfrac%7B1%7D%7B10%7DR)
And the force at this distance would be written in terms of the same equation:
Find the ratio between the final and the initial force:
Substitute the value for the final distance in terms of the initial distance:
Simplify:
This means the new force will be \frac{1}{100} of the original force.
