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elena-s [515]
3 years ago
8

Chemistry 10th grade PLEASE really need these answers :(

Chemistry
1 answer:
oksian1 [2.3K]3 years ago
7 0

Answer:

I hope this helped you.

Explanation:

A: 1

B: 3

C: 3

D: 3

E: 5

F: 5

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Which type of test helps forensic scientists to narrow down the possibilities of what drug a substance may be?
kobusy [5.1K]

The right answer is screening test.

In medicine, screening consists of looking for one or more diseases or abnormalities known as "at risk" in individuals of a given population.

These investigations are followed or not by medical consultations, clinical examinations and treatments.

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3 years ago
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Exactly 15.0 g of a substance can be dissolved in 150.0 g of water what is the solubility of the substance in grams per 100 g of
Leokris [45]
<span>(15.0 g) / (150.0 g) x (100 g) = 10.0 g/100 g H2O </span>
5 0
3 years ago
"A solution with a total volume of 850.0 mL contains 43.5 g Mg(NO3)2. If you remove 25.0 mL of this solution and then dilute thi
ycow [4]

Answer: C = 0.014M

Explanation:

From n= m/M= CV

m =43.5 M= 148, V=850ml

43.5/148= C× 0.85

C= 0.35M

Applying dilution formula

C1V1=C2V2

C1= 0.35, V1= 25ml, C2=?, V2= 600ml

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C2= 0.014M

7 0
4 years ago
One of the most difficult solar energy design problems to solve is:
Sonja [21]
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6 0
3 years ago
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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
3 years ago
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