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Dahasolnce [82]

4 years ago

10

Indicate whether each of the following indicates that a physical or chemical change has taken place when a piece of magnesium me

tal is studied: (a) Can be cut into tiny pieces (b) Fizzling occurs when placed water (c) Light is emitted when burned (d) Turns to ash

1 answer:

andrey2020 [161]4 years ago

8 0

Answer:

a) Can be cut into tiny pieces - Physical Change

b) Fizzling occurs when placed water -Chemical Change

c) Light is emitted when burned -Chemical Change

d) Turns to ash -Chemical Change

Explanation:

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Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic

Kisachek [45]

Answer:

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C) It will take 1.37 years to remove all of the arsenic from the lake.

Explanation:

A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

$\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9$

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = $164.5\times 10^{-9} g$

$1 ng=10^{-9} g$

Volume of the sample = V = $15.3 cm^3$

Density of the lake water sample ,d= $1.00 g/cm^3$

Mass of sample = M = $d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g$

$ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75$

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in $1 cm^3$ of lake water = $\frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g$

Mass of arsenic in $0.710 km^3$ lake water be m.

$1 km^3=10^{15} cm^3$

Mass of arsenic in $0.710\times 10^{15} cm^3$ lake water :

$m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g$

1 g = 0.001 kg

7,633,660.130 g = 7,633,660.130 × 0.001 kg=7,633.660130 kg ≈ 7,633.66 kg

7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

$\frac{2.74}{41.90} days$

Then days required to remove 7,633.66 kg of arsenic from the lake water :

$=7,633.66\times \frac{2.74}{41.90} days=499.19 days$

1 year = 365 days

499.19 days = $\frac{499.19}{365} years = 1.367 years\approx 1.37 years$

It will take 1.37 years to remove all of the arsenic from the lake.

3 0

3 years ago

Plz help as soon as you can

Nesterboy [21]

Answer:

98.08 g/mol

Explanation:

Hope this helps.

7 0

3 years ago

Can you guys help me answer question 5 on homogeneous mixture tysm

hjlf

Answer:

Hope this helped :) good luck! ❤️

Explanation:

A <em>coolant solution</em> is a <u><em>homogeneous </em></u>mixture because the coolant particles are not chemically combined with the water (keep their properties) and they are evenly distributed throughout the water.

7 0

3 years ago

What the correct answer

stellarik [79]

Answer:

Alkali metals are elements found in group 1 of the Periodic Table. Alkali metals are soft, malleable, ductile, and are good conductors of heat and electricity. They have low melting and boiling points compared to most other metals.

7 0

3 years ago

At a certain temperature the rate of this reaction is first order in HI with a rate constant of :0.0632s

Shalnov [3]

Answer : The time taken for the reaction is, 28 s.

Explanation :

Expression for rate law for first order kinetics is given by :

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = 0.0632

t = time taken for the process = ?

$[A_o]$ = initial amount or concentration of the reactant = 1.28 M

$[A]$ = amount or concentration left time 't' = $1.28\times \frac{17}{100}=0.2176M$

Now put all the given values in above equation, we get:

$0.0632=\frac{2.303}{t}\log\frac{1.28}{0.2176}$

$t=28s$

Therefore, the time taken for the reaction is, 28 s.

8 0

3 years ago