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MrMuchimi
3 years ago
11

When a metallic element is converted to a positive ion during a chemical reaction with oxygen, the element is said to be?

Chemistry
1 answer:
Orlov [11]3 years ago
5 0

The metal is being <em>oxidized</em>.

Metals- along with other elements- are neutral and have oxidation state of 0 in their respective elementary states.

Metal atoms can lose electrons to <em>electronegative</em> elements such as oxygen on exposure. Atoms that have lost one or more negatively-charged electrons are no longer neutral and demonstrate a positive charge. Those atoms have converted to positively charged <em>ions</em> or <em>cations</em>.

<em>Oxidation</em> is the process of <em>losing</em> electrons whereas <em>Reduction</em> refers to the process of <em>gaining</em> electrons. "OIL RIG." Metal atoms here lose electrons and are thus oxidized.

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Consider the reaction, 2 d(g) + 3 e(g) + f(g) =&gt; 2 g(g) + h(g) when h is increasing at 0.64 mol/ls, how quickly is e decreasi
bekas [8.4K]
Answer is: 1,92 mol/L·s.
Chemical reaction: 2D(g) + 3E(g) + F(g) → <span>2G(g) + H(g). 
</span>H is increasing at 0,64 mol/L·<span>s.
From chemical reaction n(H) : n(E) = 1 : 3.
0,64 mol : n(E) = 1 : 3.
n(E) = 1,92 mol.
</span>E is decreasing at 1,92 mol/L·s.
8 0
3 years ago
In an acid/base lab, a student used blue litmus paper and placed it in the sample. It stayed blue. Then the student used red lit
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Blue litmus paper in acid turns red when placed in a basic or neutral solution it remains blue. From this it can be concluded that the solution is neutral.

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3 years ago
If 20 mL of gas is subjected to a temperature change from 10°C to 100°C and a pressure change from 1 atm to 10 atm,
Evgesh-ka [11]

Answer:

E. None of these

Explanation:

We know, By GAS laws,

PV = NRT, where p- pressure, v- volume, n- number of moles, R- gas constant ,and T- temperature

Now, In the question, the number of moles remains the same as the gas is the same. so n is constant so we can compare n before and after a temperature change.

\frac{P1V1}{RT1} = \frac{P2V2}{RT2}

where P1= 1 atm, P2 = 10 atm, V1= 20 mL, T1= 10°C and T2= 100°C

We don't have to worry about the standard units as they are present equally on both the sides and get cut, same goes for R( gas constant)

So putting values, we get

\frac{1*20}{R*10} = \frac{10*V2}{R*100}

Cutting, R on both sides and moving contents to the right so that only V2 is left on the left.

\frac{1*20*100}{10*10} = V2

∴ V2 = \frac{2000}{100}

∴ V2 = 20mL

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2 years ago
Energy is gained and lost in small units or amounts called
jarptica [38.1K]
It is A. Quanta, have a nice day!!
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Answer: Vitamins

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