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Nana76 [90]
3 years ago
6

A neutral solution has a pH =

Chemistry
1 answer:
Aleonysh [2.5K]3 years ago
3 0

Answer:

7.

Explanation:

A neutral solution has a pH=7.

A basic solution has a pH>7.

An acidic solution has a pH<7.

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Balance each of the following redox reactions occurring in basic solution.MnO−4(aq)+Br−(aq)→MnO2(s)+BrO−3(aq)Express your answer
Ahat [919]

Answer : The balanced chemical equation is,

2MnO_4^-(aq)+Br^-(aq)+H_2O(l)\rightarrow 2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)

Explanation :

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 (OH^-) at that side where the less number of hydrogen are present.
  • Now balance the charge.

The half reactions in the basic solution are :

Reduction : MnO_4^-(aq)+2H_2O(l)+3e^-\rightarrow MnO_2(s)+4OH^-(aq) ......(1)

Oxidation : Br^-(aq)+6OH^-(aq)\rightarrow BrO_3^-(aq)+3H_2O(l)+6e^-  .......(2)

Now multiply the equation (1) by 2 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

2MnO_4^-(aq)+Br^-(aq)+H_2O(l)\rightarrow 2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)

8 0
3 years ago
A gas occupies a volume of 31.0 ml at 19.0°C. If the gas temperature rises to 38.0°C at constant pressure. Calculate the new vol
Alik [6]

Answer:

The correct answer is 0.0033 L (33.0 mL)

Explanation:

We uses the Charles's law which describes the changes in the volume (V) of a gas and its temperature in Kelvin (T) at constant pressure. The mathematical expression is the following:

V₁/T₁ = V₂/T₂

We have the following data:

V₁= 31.0 mL = 0.0031 L

T₁= 19.0°C = 292 K

T₂= 38.0°C = 311 K

V₂= ?

We calculate V₂ from the mathematical expression, as follows:

V₂= V₁/T₁ x T₂ = 0.0031 L/(292 K) x 311 K = 0.0033 L

5 0
3 years ago
Which of these is an isoelectronic series? 1) na+, k+, rb+, cs+ 2) k+, ca2+, or, s2– 3) na+, mg2+, s2–, cl– 4) li, be, b, c 5) n
ss7ja [257]
An isoelectronic series is where all of the ions listed have the same number of electrons in their atoms. When an atom has net charge of zero or neutral, it has equal number of protons and electrons. Hence, it means that the atomic number = no. of protons = no. of electrons. If these atoms become ions, they gain a net charge of + or -. Positive ions are cations. This means that they readily GIVE UP electrons, whereas negative ions (anions) readily ACCEPT electrons. So, to know which of these are isoelectronic, let's establish first the number of electron in a neutral atom from the periodic table:

Na=11; K=19; Rb=37; Cs = 55; Ca=20; S=16; Mg=12; Li=3; Be=4; B=5; C=6

A. Na⁺: 11-1 = 10 electrons
     K⁺: 19 - 1 = 18 electrons
     Rb⁺: 37-1 = 36 electrons

B. K⁺: 19 - 1 = 18 electrons
    Ca²⁺: 20 - 2 = 18 electrons
    S²⁻:  16 +2 = 18 electrons

C. Na⁺: 11-1 = 10 electrons
    Mg²⁺: 12 - 2 = 10 electrons
     S²⁻:  16 +2 = 18 electrons

D. Li=3 electrons
    Be=4 electrons
    B=5 electrons
    C=6 electrons

The answer is letter B.
7 0
3 years ago
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GenaCL600 [577]
Hi sure you don’t need to pay but i need what grade your in?
3 0
3 years ago
Is this atom more likely to gain electrons or to lose electrons? Explain how you can tell
elena55 [62]

Answer:

Likely to gain electrons

Explanation:

The atom shown is likely to gain additional electrons to complete its electronic configuration.

  • Since this is a neutral specie, the number of protons and electrons are the same.
  • The atom has 16 electrons
  • the number of valence electrons is 6
  • If the atom gains two additional electrons, the octet configuration is attained
  • Also, the atom can lose 6 electrons to become an octet

The atom will prefer to gain additional 2 electrons to give an octet configuration.

5 0
3 years ago
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