If the oxidation state of H is +1 and O is −2, what is the oxidation state of C in C2H4O? −1 +2 +1 −2

Answers

Answer 1

Answer: It should be noted that a neutral compound has an overall oxidation state of zero. This means that the sum of the oxidation states of each individual component multiplied by the number of atoms of that component must be equal to zero. For C2H4O, excluding the oxidation state of C, H4 gives a value of +4, and O gives a value of -2. This gives the compound a temporary oxidation value of +2. This means that in order for the compound to be neutral, C must have an oxidation state of -2. Since there are 2 atoms of C, its oxidation state must be -1.

Answer 2

Answer:

Answer:

-1

Explanation:

The sum of the oxidation states of all the atoms in a neutral compound is zero. Then the following equation must be satisfied:

number of atoms of C * oxidation state of C + number of atoms of H * oxidation state of H + number of atoms of O * oxidation state of O = 0

Replacing with data and calling the unknown x, we get:

2*x + 4*1 + 1*(-2) = 0

2*x + 2 = 0

x = (-2)/2

x = -1

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Classify the following reactions and complete the given below
A) NaCl + AgNO3 - NaNO3 + AgCl

Answers

Answer:

chemical equation balance.

Explanation:

if you need more explanation , comment.

What is the melting point of wrought iron in degrees celcius?

Answers

The melting point of iron is 2,800°F or 1,538°C.

the radioisotope radon-222 has a half-life of 3.8 days. How much of a 73.9-gram sample of radon-222 would be left after approximately 23 days?

Answers

There will be 1.11 grams of the sample of radon-222 left. This answer can be obtained using the formula of half life to get the rate constant which will be used in another equation later on.

Half-life (t) = ln2/k = 3.8 days

k = 0.182407/day

Using the general equation of a first order reaction:

Ca/Cao = 1/e^(kt)
Ca/Cao = 0.01506 --> fraction of radon-222 left

This means that 1.51% of the original amount remains, so 1.51% of 73.9 is 1.11 grams.

Forces and Motion:Question 2A student pushes a box to slide it across the floor. Which statement best explains why the box begins
to move?
Select one:
0 The weight of the box is less than the force of friction.
O
The weight of the box is greater than the force of friction.
The applied force is greater than the force of friction.
O
The applied force is less than the force of friction

Answers

Answer:

the applied force is less than the force of friction.

Explanation:

You discover a new type of gland associated with the skin. chemical analysis of the product shows a secretion has a ph of 4, consists of 99% water, and contains traces of normal electrolytes including urea, vitamin c, and dermicidin. there are no traces of fats or proteins. how would you classify this new gland?

Answers

This new gland would most likely be an eccrine gland, meaning it would be a sweat gland. They are found all over the body and can also be called merocrine glands.

This new gland would most likely be an eccrine gland, meaning it would be a sweat gland.
They are found all over the body and can also be called merocrine glands.

Explanation:
The 2 main forms of sweat glands are eccrine sweat glands and apocrine sweat glands.

Eccrine sweat glands are smaller sweat glands. they're volute cannular glands that discharge their secretions onto the surface of the skin sweat gland.
any of the rather tiny secreter that manufactures a fluid secretion while not removing protoplasm from the secreting cells which are restricted to the human skin is called eccrine sweat gland.

How many grams of Cl are in 535 g of CaCl2?

Answers

m( $Ca Cl_(2)$ )=535 g
m(Cl)=?
The molar mass of $CaCl_(2)$ = 40.08 + 2 * 35.45 = 110.98
The atomic mass of Cl: 35.45 ( you can see it in the Periodic table)
110.98 g $CaCl_(2)$ - 35.45 * 2 g Cl
535 g $CaCl_(2)$ - x g Cl
Proportion:
110.98 : 535 = (35,45 * 2) : x
x = 535 * 35.45 * 2 / 110.98 = 37,931.5/110.98 = 341.79 g
Answer: m(Cl)=341.79 g

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