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4 years ago

What element has a mass number of 16 and 8 neutrons

2 answers:

8 0

The element would be Oxygen.
Work:
Mass = atomic # + neutrons
(The atomic number is the same as the number of protons in an element)

To find the number of protons substitute x
16 = x + 8
x = 8 protons
Oxygen has 8 protons

I hope this helps!

Simora [160]4 years ago

3 0

The answer is O (oxygen).

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How many moles of NaCl (s) can be formed from 32 moles of Cl2 (g) reacting with an excess of Na (s)?

blagie [28]

Answer:

64 moles

Explanation:

Let us begin by writing a balanced equation for the reaction. This is illustrated below:

2Na + Cl2 —> 2NaCl

From the equation above,

1 mole of Cl produced 2 moles of NaCl.

Therefore, 32 moles of Cl will produce = 32 x 2 = 64 moles of NaCl.

Therefore, 64 moles of NaCl are produced

7 0

4 years ago

Answer choices Metals Metalloids nonmetals

lbvjy [14]

Answer:

nonmetals

Explanation:

the answer is nonmetals

7 0

3 years ago

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What mass, in grams, of bromine gas (Br2) is contained in a 15.7-liter tank at 24.6 degrees Celsius and 0.986 atmospheres? Show

Nezavi [6.7K]

PV = nRT

P is pressure, V is volume, n is number of moles, R is the gas constant, T is temperature in K

(2.85 atm)(12.5 L) = (n)(.08206)(27 C + 273)

n = 1.45 moles x 35.45 grams / mol Cl2 = 51.3 grams

4 0

3 years ago

Read 2 more answers

A square block of steel with volume 10 cm3 and mass of 75 g is cut precisely in half the density of the two smaller pieces is no

Roman55 [17]

The density of a substance is its mass per unit volume, given by the mass divided by the volume. Density depends on the substance of concern and doesn't matter whether one has a little amount of it or a whole full of it. Therefore, the density of the two smaller pieces is now the same as the original density.

3 0

3 years ago

2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electrons

Julli [10]

Answer :

The oxidation state of tin changes from (+2) to (+4).

The number electrons transferred in the reaction are, 10 electron.

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in basic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion $(H^+)$ at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

$2BrO_3^-+5SnO_2^{2-}+H_2O\rightarrow 5SnO_3^{2-}+Br_2+2OH^-$

The oxidation-reduction half reaction will be :

Oxidation : $SnO_2^{2-}\rightarrow SnO_3^{2-}$

Reduction : $BrO_3^-\rightarrow Br_2$

First balance the main element in the reaction.

Oxidation : $SnO_2^{2-}\rightarrow SnO_3^{2-}$

Reduction : $2BrO_3^-\rightarrow Br_2$

Now balance oxygen atom on both side.

Oxidation : $SnO_2^{2-}\rightarrow SnO_3^{2-}+H_2O$

Reduction : $2BrO_3^-+H_2O\rightarrow Br_2$

Now balance hydrogen atom on both side.

Oxidation : $SnO_2^{2-}+OH^-\rightarrow SnO_3^{2-}+H_2O$

Reduction : $2BrO_3^-+3H_2O\rightarrow Br_2+12OH^-$

Now balance the charge.

Oxidation : $SnO_2^{2-}+OH^-\rightarrow SnO_3^{2-}+H_2O+2e^-$

Reduction : $2BrO_3^-+3H_2O+10e^-\rightarrow Br_2+12OH^-$

The charges are not balance on both side of the reaction. We are multiplying oxidation reaction by 5 and then added both equation, we get the balanced redox reaction.

Oxidation : $5SnO_2^{2-}+5OH^-\rightarrow 5SnO_3^{2-}+5H_2O+10e^-$

Reduction : $2BrO_3^-+3H_2O+10e^-\rightarrow Br_2+12OH^-$

The balanced chemical equation in basic medium will be,

$2BrO_3^-+5SnO_2^{2-}+H_2O\rightarrow 5SnO_3^{2-}+Br_2+2OH^-$

The number electrons transferred in the reaction are, 10 electron.

7 0

3 years ago