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evablogger [386]

3 years ago

6

A chemistry instructor provides each student with 8 test tubes at the beginning of the year. If there are 28 students per class,

how many test tubes are required for three chemistry classes?

1 answer:

klemol [59]3 years ago

7 0

Three classes: 28×3=84 students
8 test tubes per student: 84× 8= 672
hope this helps!

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A geochemist measures the concentration of salt dissolved in Lake Parsons, an isolated salt lake. He finds a concentration of 21

VikaD [51]

Answer:

68,2%

Explanation:

Supposing the initial salt concentration of lake Parsons is the same of non-isolated lakes, 6,67L, and the change of salt concentration in isolated lake is just for water evaporation it is possible to write:

6,67gL⁻¹×X = 21gL⁻¹×Y

<em>-Where X is the initial water and Y is the water that remains in the isolated lake-</em>

Thus:

6,67X = 21Y

0,318 = Y/X

0,318 is the ratio of water that remains between total water. To obtain the ratio of evaporated water:

1-0,318 = 0,682

In percentage: <em>68,2%</em>

<em />

I hope it helps!

<em />

3 0

3 years ago

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,

Masja [62]

<u>Answer:</u> The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

<u>Explanation:</u>

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

<u>Oxidation half reaction:</u> $H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V$ ( × 3)

<u>Reduction half reaction:</u> $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

<u>Net reaction:</u> $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

Hence, the concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

7 0

3 years ago

Which of the following is the best example of a scientific law?

il63 [147K]

Answer:

The correct answer is :'The gravitational force between two objects is related mathematically to their mass and the distance between them'.

Explanation:

A scientific law is a statement which explains physical phenomena and has been tested multiple times through the correct scientific methods and experimentation. It will give same results and observations after each conduct.

A scientific law comes into picture after the acceptance of a theory which explains or exposes the reason behind any physical phenomena occurrence.

So the best example of scientific law:

The gravitational force between two objects is related mathematically to their mass and the distance between them.

5 0

3 years ago

Read 2 more answers

Hunting lions has been banned in some parts of Africa. If this law is enforced, what can we expect to happen to the populations

k0ka [10]

The population would decrease, the more predators there are the more food needed for the species .

4 0

3 years ago

Read 2 more answers

Part A Which acid in each of the following pairs has the stronger conjugate base? Match the words in the left column to the appr

vichka [17]

Answer:

HF

H₂S

H₂CO₃

NH₄⁺

Explanation:

<em>Which acid in each of the following pairs has the stronger conjugate base?</em>

According to Bronsted-Lowry acid-base theory, <em>the weaker an acid, the stronger its conjugate acid</em>. Especially for weak acids, pKa gives information about the strength of such acid. <em>The higher the pKa, the weaker the acid.</em>

<em />

Of the acids HCl or HF, the one with the stronger conjugate base is HF because it is a weak acid.

Of the acids H₂S or HNO₂, the one with the stronger conjugate base is H₂S because it is a weaker acid. pKa (H₂S) = 7.04 > pKa (HNO₂) = 3.39

Of the acids H₂CO₃ or HClO₄, the one with the stronger conjugate base is H₂CO₃ because it is a weak acid.

Of the acids HF or NH₄⁺, the one with the stronger conjugate base is NH₄⁺ because it is a weaker acid. pKa (HF) = 3.17 < pKa (NH₄⁺) = 9.25

6 0

3 years ago