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3 years ago

What do navigators need to consider when plotting a course?

1 answer:

8 0

1.) scale of the chart
2.) Notes of the chart
3.) chart symbols
4.) chart corrections
5.) GPS positions
6.) radar fixes

7.) Visual fixtures and position circle and position line

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C. What happens to stuff when you add heat energy to it?

Minchanka [31]

Answer:

it warms up

Explanation:

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3 years ago

Read 2 more answers

How many moles of oxygen are needed to react with 7.5 moles of aluminum?

LuckyWell [14K]

Equation for oxygen gas molecules
1.5O₂ + 2Al => Al₂O₃
ratio of aluminium to oxygen is
2:1.5
when Al is 7.5
O₂ is 3.75

4 0

4 years ago

A simple reaction: 2A+B=C. IF THE CONCENTRATION OF B IS INCREASED 4 TIMES,HOW MANY TIMES SHOULD THE CONCENTRATION OF A DECREASE

hichkok12 [17]

Answer:

is the length of floor plan with an area of 8q and how to use the pool to use the pool to use the pool to use the pool to to use the the pool to use to grind a grind and with a diameter of 3m and a ads a see if this is a is possible required for a quiz night or a going up with a diameter question to for Filipinos nga ba eh nga ba eh nga maganda sa at ang the night when we get the chance to watch it ⌚ and it's just the first time six 6⃣ and we have all been the best proud of of the music that music has ever made it a good

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3 years ago

Use the Henderson-Haaselbach equation based on the buffer equilibrium reaction to complete the statements below.

zloy xaker [14]

Hi, Again!

The answer to the drop downs are

•Increase
•Decreases
• 1:1
•5.9

Hope this helps !

8 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