Which of the following is a product in the reaction described by the word equation below?

Where can you find most of Earth's liquid freshwater?

KengaRu [80]

Answer:

Where can you find most of Earth's liquid freshwater?

Explanation:

Over 68 percent of the fresh water on Earth is found in icecaps and glaciers, and just over 30 percent is found in ground water. Only about 0.3 percent of our fresh water is found in the surface water of lakes, rivers, and swamps.

Hope that helped, i also have snap if you need me.

6 0

3 years ago

If a radio station transmits on AM 610, how many hertz (Hz) is the frequency of the wave? (Remember that kHz = kilohertz.)

elixir [45]

Answer:

610,000

Explanation:

8 0

3 years ago

The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.

enot [183]

Answer:

$\large \boxed{\text{21.6 L}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 24.5

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}$

(b) Moles of CO₂

$\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}$

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37 °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

$\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}$

7 0

2 years ago

How do you convert 134kj to Calories?

Sonja [21]

1 kilo joule = 0.239006 calories
134 kilo joule = 134 x 0.239006
= 32.026804 calories

5 0

3 years ago

A 4.86-gram sample of calcium reacted completely with oxygen to form 6.80 grams of calcium oxide. This reaction is represented b

sesenic [268]

The number of mole of Ca reacted is:
4.86 g Ca/ (40.08 g/mol Ca)= 0.121 mol Ca

Because Ca reacted completely with oxygen and there is 2 mol Ca, there is 1 mol O2 reacted.

Total mass of oxygen that reacted is:
0.121 mol Ca* (1mol O2/ 2 mol Ca)* (32 g O2/ 1 mol O2)= 1.94 g O2 reacted.

Hope this would help~

8 0

3 years ago