All categories

3 years ago

Please help illl mark brainlst (no links)

Chemistry

1 answer:

Arte-miy333 [17]3 years ago

6 0

Answer:

A ) New rocks floating to the surface of the crust.

TRUST

You might be interested in

Calculate the number of molecules in 0.75 moles of CH₄. x 10²³ molecules

spin [16.1K]

Ans :4.5 * 10²³ molecules

Given:

Moles of CH4 = 0.75

To determine;

The number of molecules of CH4 in 0.75 moles

Explanation:

1 mole of CH4 contains Avogadro's number i.e. 6.023*10²³ molecules

Therefore, 0.75 moles would correspond to :

= 0.75 moles * 6.023*10²³ molecules/1 mole

= 4.5 * 10²³ molecules

7 0

3 years ago

Read the following story and then answer the questions.

Sliva [168]

Place both breaking in a freezer at a temperature

4 0

3 years ago

Read 2 more answers

You have a solution that is 18.5% (v/v) methyl alcohol. If the bottle contains 1.64 L of solution, what is the volume (V) in mil

Juliette [100K]

Answer:

303.4 millimeters

Explanation:

Just multiply 1.64 by 0.185. then convert that answer to millimeters

7 0

2 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

Why must we do the a lot of quantity urine

serg [7]

Answer:

because

ExplanatioN:

BeCaUsE

7 0

3 years ago