All categories

3 years ago

Are the changes caused by wind erosion of land in the high plains all due to human activity?

Chemistry

1 answer:

Feliz [49]3 years ago

4 0

No wind is natural. Need to write something to make it 20 letters :)

You might be interested in

Which of the following is an example of a buoyant force acting on a piece of

hodyreva [135]

Answer:

D. The piece of bread travels with straight- line horizontal acceleration.

Explanation

4 0

3 years ago

what is the consequence on your brain for snorting or taking amphetamines(meth/stimulant) as a 16 year old(minor)?

Yanka [14]

There are many consequences!

Some being:

* ADDICTION

* HEALTH ISSUES

* MENTAL ISSUES

5 0

2 years ago

Read 2 more answers

Suggest why it might be difficult and dangerous to test sodium for electrical conductivity?

lukranit [14]

Answer:

It might cause fire or extreme smoke.

Explanation:

8 0

2 years ago

What is the percent composition by mass for hydrogen in water (H2O)?

saw5 [17]

Explanation:

Percentage composition = 2/18 = 11.11%.

5 0

3 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

7 0

3 years ago