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Artist 52 [7]
2 years ago
11

What features of this model will help Armando answer the question?

Chemistry
1 answer:
ASHA 777 [7]2 years ago
5 0

Answer:

Explanation:

Answer to Armando is an artist who sells prints of his original paintings. ... This problem has been solved! ... Armando wants to create a function P(x). to model his total profit where x is the number of paintings sold

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Energy is just like any other chemical or physical process, where it cannot be created or destroyed. Which Law states this fact?
Ganezh [65]

Answer:

Law of Conservation of Energy

Explanation:

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I need help with this
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Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reactionP4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°r
weqwewe [10]

Answer:

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

Explanation:

Enthalpy is denoted by H.

Enthalpy: Total heat change in a chemical reaction is called enthalpy.

The change of entalpy of a reaction is denoted by \bigtriangledown H^\circ_{rxn}

Hass's Law:The change in enthalpy of any process can be determined by calculating the sum of change in enthalpy of each of the steps involved in the process.

g= gas

S= solid

P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}=?

PCl₅(s)→ PCl₃(g)+Cl₂(g) .......(1)       \bigtriangledown H^\circ_{rxn}= +157KJ

P₄(g)+6Cl₂(g)→  4PCl₃(g).............(2)     \bigtriangledown H^\circ_{rxn}= -1207 KJ

If we flip a reaction the value of enthalpy will be change positive to negative or nagative to positive but the numerical value will be remain same.

We need rearrange the equation (1) because in the required equation Cl₂ is on the left side. So we flip the first equation.

PCl₃(g)+Cl₂(g)→PCl₅(s)......(3)          \bigtriangledown H^\circ_{rxn}= -157KJ

Multiplying 4 with equation (3)

4 PCl₃(g)+4Cl₂(g)→4PCl₅(s)......(4)          \bigtriangledown H^\circ_{rxn}=4×( -157)KJ= -628 KJ

Adding equation (2) and (4) we get

P₄(g)+6Cl₂(g)+4 PCl₃(g)+4Cl₂(g)→4PCl₃(g)+4PCl₅(s)    \bigtriangledown H^\circ_{rxn}=( -1207-628)KJ

⇒P₄(g)+10Cl₂(g)→4PCl₃(g)-4PCl₃(g)+4PCl₅(s)      \bigtriangledown H^\circ_{rxn}= - 1835KJ

⇒P₄(g)+10Cl₂(g)→ 4Cl₅(s)       \bigtriangledown H^\circ_{rxn}= -1835 KJ

Therefore  \bigtriangledown H^\circ_{rxn}= -1835 KJ

5 0
2 years ago
Which element has a complete valence electron shell?
borishaifa [10]
Argon, it's a noble gas in the last group on the periodic table.
7 0
3 years ago
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If u (x) = negative 2 x squared and v (x) = StartFraction 1 Over x EndFraction, what is the range of (u circle ma poottay
KonstantinChe [14]
The range is negative numbers.
The interval for the range is .
***You might want to look at your functions again because I don't see a choice that matches.
Step-by-step explanation:
Given functions:


We are asked to find the range of .
I'm also going to look at the domain just to see if this possibly might change my range .
is the inner function. So we will consider the domain of that function first.
You only have to worry about division by zero for the function .
Since we are dividing by , we don't want to be zero.
So far the domain is all real numbers except .
Now let's move out.
exists for all numbers, . So we didn't want to include from before.
Now let's put it together:







So the domain is still all real numbers except at since we cannot divide by 0 and is 0 when .
with .
is positive for all numbers except .
So is negative for all numbers since negative divided by positive is negative.
So the range is only negative numbers.
Let's also look at the inverse:

Multiply both sides by :

Divide both sides by :

Take the square root of both sides:
.
So can't be 0 and it also can't be positive because the inside of the square root will be negative (since negative divided by positive results in negative).
5 0
3 years ago
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