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3 years ago

What is the difference between a rock and a mineral?

Chemistry

1 answer:

SVEN [57.7K]3 years ago

5 0

Answer:

A rock is an inconsistent mixture of minerals, while mineral is "pure" and made up of a precise combination of chemicals.

Explanation:

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A sample of N2 gas in a flask is heated from 27 Celcius to 150 Celcius. If the original gas is @ pressure of 1520 torr, what is

Romashka [77]

Answer:

$\large \boxed{\text{B.) 2.8 atm}}$

Explanation:

The volume and amount are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

$\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}$

Data:

p₁ = 1520 Torr; T₁ = 27 °C

p₂ = ?; T₂ = 150 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = ( 27 + 273.15) K = 300.15 K

T₂ = (150 + 273.15) K = 423.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{1520}{300.15} & = & \dfrac{p_{2}}{423.15}\\\\5.064 & = & \dfrac{p_{2}}{423.15}\\\\5.064\times423.15&=&p_{2}\\p_{2} & = & \text{2143 Torr}\end{array}\\$

(c) Convert the pressure to atmospheres

$p = \text{2143 Torr} \times \dfrac{\text{1 atm}}{\text{760 Torr}} = \textbf{2.8 atm}\\\\\text{The new pressure reading will be $\large \boxed{\textbf{2.8 atm}}$}$

7 0

3 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

4 years ago

One mole of carbon, sulfur, and aluminum, respectively. Which jar contains the most atoms?

forsale [732]

Answer:

sulphur cuz its weight is 32.065g

Explanation:

this is correct

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7 0

3 years ago

Read 2 more answers

Which structure is part of the central nervous system?

Oxana [17]

The spinal cord is part of the central nervous system

4 0

3 years ago

Measurements show that unknown compound has the following composition: element mass 38.7 % calcium, 19.9 % phosphorus and 41.2 %

Mademuasel [1]

Answer:

D) empirical formula is: C₃P₂O₈

Explanation:

Given:

Mass % Calcium (Ca) = 38.7%

Mass % Phosphorus (P) = 19.9%

Mass % oxygen (O) = 41.2 %

This implies that for a 100 g sample of the unknown compound:

Mass Ca = 38.7 g

Mass P = 19.9 g

Mass O = 41.2 g

Step 1: Calculate the moles of Ca, P, O

Atomic mass Ca = 40.08 g/mol

Atomic mass P = 30.97 g/mol

Atomic mass O = 16.00 g/mol

$Moles\ Ca = \frac{38.7g}{40.08g/mol} =0.966\ mol\\\\Moles\ P = \frac{19.9g}{30.97g/mol} =0.643\ mol\\\\Moles\ O = \frac{41.2g}{16.00g/mol} =2.58\ mol$

Step 2: Calculate the molar ratio

$C = \frac{0.966}{0.643} =1.50\\\\P = \frac{0.643}{0.643} = 1.00\\\\O = \frac{2.58}{0.643} =4.00$

Step 3: Calculate the closest whole number ratio

C: P: O = 1.50 : 1.00 : 4.00

C : P : O = 3:2:8

Therefore, the empirical formula is: C₃P₂O₈

7 0

3 years ago