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dezoksy [38]
3 years ago
8

PLEASE HELP ME YOU'LL GET JUST 10 POINTS.

Chemistry
1 answer:
Sphinxa [80]3 years ago
4 0
I want more points for this answer
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1. Choose the correct answer from the given alternative. 1. A blood sample analysis showed that a total of 0.0000036g lead. Expr
son4ous [18]

Answer:

3.6 micrograms answers (c)

Explanation:

5 0
2 years ago
Answer truthfully:))​
Elina [12.6K]

Answer:

<u>Formula</u><u>:</u> Velocity \:  V = f \lambda \\  Solution: = 5 \times 0.8 \\  = 4 \:  {ms}^{ - 1}

3 0
2 years ago
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1) Consider this row in the periodic table. What changes can you predict based on what information is provided by the boxes for
bearhunter [10]

Answer:

B) As you move across the row, the number of electrons increases and reactivity also increases.

Explanation:

The periodic table is arranged in a way that if you go across a period, the number of protons, neutrons, and electrons in an element increases. In terms of reactivity, the most reactive elements are the ones which have a high electronegativity. The electronegativity of the elements increases as you travel to the right and upwards on the periodic table.

4 0
2 years ago
Al(s) + NiSO4(aq)<br> Balance
irinina [24]

Answer:

2Al(s) + 3NiSO4 --> Al2(SO4)3 + 3Ni

Explanation:

This reaction type is a single replacement. The format of a single replacement is:

A+BC --> B+AC

A= Al

B= Ni

C= SO

The coefficient 3 for Ni would become a subscript for  AC. After you plug those into the reaction you need to count how many of each are on the left side and try to get the same number on the right side. Both sides must be equal to have a balanced equation.

8 0
3 years ago
Consider the following reaction:
iren [92.7K]

Answer:

A. ΔG° = 132.5 kJ

B. ΔG° = 13.69 kJ

C. ΔG° = -58.59 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpy of formation

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))

ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)

ΔH° = 178.3 kJ

We can calculate the standard entropy of the reaction (ΔS°) using the following expression.

ΔS° = ∑np . S°p - ∑nr . S°r

where,

S: standard entropy

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))

ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)

ΔS° = 160.6 J/K. = 0.1606 kJ/K.

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

T: absolute temperature

<h3>A. 285 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ

<h3>B. 1025 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ

<h3>C. 1475 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ

5 0
3 years ago
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