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earnstyle [38]
3 years ago
5

Plz help! This question is due in 10 mins!

Chemistry
1 answer:
Gnom [1K]3 years ago
7 0

Answer: 20

Explanation: I’m in your class.

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Look at the following diagram of the carbon cycle.
katrin2010 [14]

Answer:

The energy consumed by animals in the form of glucose is conserved because it is transformed into chemical energy as carbon dioxide is produced during respiration.

Explanation:

There's no diagram....but I kinda figured it from the description.

4 0
3 years ago
How many liters of a 0.5 m sodium hydroxide solution would contain 2 moles of solute?
Klio2033 [76]

Answer:

  • 4 liters

Explanation:

<u>1) Data:</u>

a) M = 0.5 M

<u>Note</u>: since the question is about volume (liters) and moles of solute, the correct unit for the concentration of the hydroxide solution is M (upper case) which is the unit for molarity. Lower case m is used for molality, and you cannot calculate the volume of the solution unless you know some other data, e.g. density of the solution.

b) n = 2

c) V (in liters) = ?

<u>2) Formula:</u>

Molarity is the number of moles of solute per liter of solution:

  • M = n / V (in liters)

<u>3) Solution:</u>

a) Clear V from the formula: V (in liters) = M / n

b) Substitute: V (in liters) =  2 moles /  0.5M  = 4 liters ← answer

7 0
3 years ago
9) What is the density of the gold rock with a mass of 386 grams and has a volume of 20 cubic centimeters?
miss Akunina [59]

Answer:

density= mass/volume

hence density of gold rock

= 386/20

=19.3 g/cc

3 0
3 years ago
Which scientist did people think was insane so they dismissed his ideas for almost 2000 years?
forsale [732]
Alfred Wegener


I think let me know if I am wrong
6 0
3 years ago
Muriatic acid, an industrial grade of concentrated HCl, is used to clean masonry and cement. Its concentration is 11.7 M. For ro
Nata [24]

Answer:

\boxed{\text{199 mL}}

Explanation:

1. Calculate the moles of HCl  

n = \text{25.4 g} \times \dfrac{\text{1 mol}}{\text{36.46 g}} = \text{0.6967 mol}

2. Calculate the volume of dilute HCl

\begin{array}{rcl}\text{Molar concentration}& =& \dfrac{\text{moles}}{\text{litres}}\\\\3.50 & = & \dfrac{0.6967}{V}\\\\3.50V & = & 0.6967\\\\V & = & \dfrac{0.6967}{3.50}\\\\V & = & \text{0.199 L} =\textbf{199 mL}\end{array}\\\text{The volume of muriatic acid needed is }\boxed{\textbf{199 mL}}

3 0
3 years ago
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