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Alex73 [517]
3 years ago
11

An atom of a certain element has 19 protons, 19 electrons, and a mass number of 39. This element is very reactive. How many neut

rons are in this atom?
Chemistry
1 answer:
gogolik [260]3 years ago
6 0

Answer:

There would be 20 neutrons in that atom.

Explanation:

The mass number of an atom is the sum of:

  • the number of neutrons in this atom, and
  • the number of protons in this atom.

Electrons do not contribute to the mass number of an atom.

From the question:

  • Mass number of this atom: 39.
  • Number of protons in this atom: 19.

(\text{Mass Number}) = (\text{Number of protons}) + (\text{Number of neutrons}).

In this atom:

39 = 19 + (\text{Number of neutrons}).

Hence, the number of neutrons would be 39 - 19 = 20.

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0.0145 moles of helium gas are introduced into a balloon so that the volume of the balloon is 2.54 liters. An additional amount
olga_2 [115]

Answer:

4.43L is final volume of the ballon

Explanation:

Avogadro's law of ideal gases states that <em>equal volumes of gases, at the same temperature and pressure, have the same number of molecules</em>.

The formula is:

\frac{V_1}{n_1} =\frac{V_2}{n_2}

Where V and n are volume and moles of the gas in initial and final conditions.

If the initial conditions are 0.0145 moles and 2.54L and final amount of moles is 0.0253moles, final volume is:

\frac{2.54L}{0.0145mol} =\frac{V_2}{0.0253mol}

V₂ = <em>4.43L is final volume of the ballon</em>

6 0
3 years ago
Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial conc
Anni [7]

Answer:

Ka = 4.76108

Explanation:

  • CO(g) + 2H2(g) ↔ CH3OH(g)

∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]

                      [ ]initial         change         [ ]eq

CO(g)              0.27 M       0.27 - x        0.27 - x

H2(g)              0.49 M       0.49 - x        0.49 - x

CH3OH(g)          0                0 + x               x = 0.11 M

replacing in Ka:

⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)

⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)

⇒ Ka = (0.11) / (0.38)²(0.16)

⇒ Ka = 4.76108

7 0
3 years ago
Draw the structure of tofacitinib with explicitly shown all carbon and hydogen atoms. identify the number of carbon atoms in tof
Andrew [12]

The structure of tofacitinib with explicit carbon and hydrogen is the first attached picture.

The number of carbon atoms in tofacitinib are 16.

The molecular formula of tofacitinib is C_{16}H_{20}N_{6}O.

The structure of tofacitinib with all the filled hydrogen atoms is the second attached picture.

3 0
3 years ago
Write the overall, balanced molecular equation and indicate which element is oxidized and which is reduced for the following rea
REY [17]

Explanation:

Mg(s) + Cr(C2H3O2)3 (aq)

Overall, balanced molecular equation

Mg(s) + Cr(C2H3O2)3(aq) --> Mg(C2H3O2)3(aq) + Cr(s)

To identify if an element has been reduced or oxidized, the oxidation number is observed in both the reactant and product phase.

An increase in oxidation number denotes that the element has been oxidized.

A decrease in oxidation number denotes that the element has been reduced.

Oxidation number of Mg:

Reactant - 0

Product - +3

Oxidation number of Cr:

Reactant - +3

Product - 0

Note: C2H3O2 is actually acetate ion; CH3COO- The oxidatioon number of C, H and O do not change.

Oxidized : Mg

Reduced : Cr

7 0
3 years ago
6) A volume of 473 mL of oxygen was collected at 27°C. What volume would the oxygen occupy at
sertanlavr [38]

Answer : The volume of oxygen occupy at 173° would be, 703.2 mL

Explanation :

Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

where,

V_1\text{ and }T_1 are the initial volume and temperature of the gas.

V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=473mL\\T_1=27^oC=(27+273)K=300K\\V_2=?\\T_2=173^oC=(173+273)K=446K

Now put all the given values in above equation, we get:

\frac{473mL}{300K}=\frac{V_2}{446K}\\\\V_2=703.2mL

Therefore, the volume of oxygen occupy at 173° would be, 703.2 mL

6 0
2 years ago
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