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Darina [25.2K]
3 years ago
8

What is the molar mass of c25h52​

Chemistry
2 answers:
Cloud [144]3 years ago
8 0
The molar mass of c25h52 is 352.77g/mol
harkovskaia [24]3 years ago
6 0
The molar mass is 352.77g/mol
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What is the mass of 4 moles of almunium atom​
lana [24]

Answer:

108 grams

Explanation:

5 0
2 years ago
Part 1: Cats falling out of windows
timurjin [86]

Answer:

A cat that falls out of the window even if it seems to be doing well, must urgently go to a veterinary clinic in order to be placed under surveillance for at least 24 to 72 hours

Explanation:

3 0
3 years ago
Calcium cyanamide, CaCN2, reacts with water to form calcium carbonate and ammonia. CaCN2(s)+3H2O(l)→CaCO3(s)+2NH3(g). How many g
Norma-Jean [14]

298 g of calcium carbonate CaCO₃

Explanation:

We have the following chemical reaction:

CaCN₂ (s) + 3 H₂O (l) → CaCO₃ (s)+ 2 NH₃ (g)

number of moles = mass / molar weight

number of moles of H₂O = 161 / 18 = 8.94 moles

Knowing the chemical reaction we devise the following reasoning:

if        3 moles of H₂O produces 1 mole of CaCO₃

then  8.94 moles of H₂O produces X moles of CaCO₃

X = (8.94 × 1) / 3 = 2.98 moles of CaCO₃

mass =  number of moles × molar weight

mass of CaCO₃ = 2.98 × 100 = 298 g

Learn more about:

number of moles

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7 0
3 years ago
What volume of air at 25°C and 1.00 atm can he stored in a 10.0 L high-pressure air tank if compressed to 25°C and 175 atm?
DaniilM [7]

Answer:

1750L

Explanation:

Given

Initial Temperature = 25°C

Initial Pressure = 175 atm

Initial Volume = 10.0L

Final Temperature = 25°C

Final Pressure = 1 atm

Final Volume = ?

This question is an illustration of ideal gas law.

From the given parameters, the initial temperature and final temperature are the same; this implies that the system has a constant temperature.

As such, we'll make use of Boyle's Law to solve this;

Boyle's Law States that:

P₁V₁ = P₂V₂

Where P₁ and P₂ represent Initial and Final Pressure, respectively

While V₁ and V₂ represent Initial and final volume

The equation becomes

175 atm * 10L = 1 atm * V₂

1750 atm L = 1 atm * V₂

1750 L = V₂

Hence, the final volume that can be stored is 1750L

5 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
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