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gizmo_the_mogwai [7]
2 years ago
10

What is the name of this compound Fe(NH3)5Cl4​

Chemistry
1 answer:
aev [14]2 years ago
3 0

Answer:

it a complex salt.......

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A solution containing a nonvolatile solute will have
svetoff [14.1K]
It is nonvolatile solute
5 0
2 years ago
When a small amount of 12 M HNO3(aq) is added to a buffer solution made by mixing CH3NH2(aq) and CH3NH3Cl(aq) , the pH of the bu
fredd [130]

Answer:

a. CH3NH2(aq) + H⁺ → CH3NH3⁺

Explanation:

The mixture of a weak base as CH3NH2 with its conjugate acid CH3NH3Cl produce a buffer. As the weak acid is in equilibrium with water, the mixture of the weak base and its conjugate base produce that the acid or base released react avoiding the change in pH.

For example, when a strong acid as HNO3 reacts, the weak base will react producing the conjugate base, that is:

CH3NH2(aq) + H⁺ → CH3NH3⁺

Right answer is:

<h3>a. CH3NH2(aq) + H⁺ → CH3NH3⁺</h3>

5 0
2 years ago
If I have one mole of sulfur, how many atoms would that be?
krok68 [10]

Answer:

Atoms of sulfur = 9.60⋅g32.06⋅g⋅mol−1×6.022×1023⋅mol−1

Explanation:

because the units all cancel out, the answer is clearly a number, ≅2×1023 as required.

7 0
3 years ago
An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +
Maurinko [17]

Answer:

The simplified expression for the fraction  is  \text {X} =    \dfrac{  {k_3  \times cM} }{k_1 +k_2 + k_3 }

Explanation:

From the given information:

O3* → O3                   (1)    fluorescence

O + O2                      (2)    decomposition

O3* + M → O3 + M    (3)     deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

\text {X} =    \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) }  } }

\text {X} =    \dfrac{  {k_3 \times cO \times cM} }{  {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) }  }

\text {X} =    \dfrac{  {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3  \times cM) }

\text {X} =    \dfrac{  {k_3  \times cM} }{k_1 +k_2 + k_3  }    since  cM is the concentration of the inert molecule

7 0
3 years ago
If Log 4 (x) = 12, then log 2 (x / 4) is equal to
Alexus [3.1K]

The value of log₂(x/4) is 22. Using the properties of the logarithm, the required value is calculated.

<h3>What are the required properties of the logarithm?</h3>

The required logarithm properties are

logₐx = n ⇒ aⁿ = x; and logₐ(xⁿ) = n logₐ(x);

Where a is the base of the logarithm.

<h3>Calculation:</h3>

It is given that,

log₄(x) = 12;

On applying the property logₐx = n ⇒ aⁿ = x; here a = 4;

So,

log₄(x) = 12 ⇒ 4¹² = x

⇒ x = (2²)¹² = 2²⁴

Then, calculating log₂(x/4):

log₂(x/4) = log₂(2²⁴/4)

              = log₂(2²⁴/2²)

              = log₂(2²⁴ ⁻ ²)

              = log₂(2²²)

On applying the property logₐ(xⁿ) = n logₐ(x);

log₂(x/4) = 22 log₂2

We know that logₐa = 1;

So,

log₂(x/4) = 22(1)

∴ log₂(x/4) = 22.

Learn more about the properties of logarithm here:

brainly.com/question/12049968

#SPJ9

8 0
1 year ago
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