All categories

3 years ago

1. At STP, you have 0.101 g of carbon dioxide gas. What is the volume of this gas?

Chemistry

1 answer:

Ber [7]3 years ago

3 0

Answer:

1) A. 0.0515 L

2) B. 35.504 L

3) B. 422.446 g

Explanation:

At STP, you have 0.101 g of carbon dioxide gas. What is the volume of this gas?

Step 1: Data given

Mass of CO2 gas = 0.101 grams

Molar mass of CO2 = 44.01 g/mol

STP = 1 atm and 273K

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 0.101 grams / 44.01 g/mol

Moles CO2 = 0.0023 moles

Step 3: Calculate the volume

For 1 mol at STP we have 22.4L

0.0023 moles we have 22.4L * 0.0023 = 0.0515 L

Option A is correct

You have 27g of ammonia (nitrogen trihydride) at STP. What is the volume? Hint...you will first need to convert mass of ammonia to moles!

Step 1: Data given

Molar mass of NH3 = 17.03 g/mol

Mass of NH3 = 27.00 grams

STP = 1 atm and 273K

Step 2: Calculate moles NH3

Moles NH3 = 27.00 grams / 17.02 g/mol

Moles NH3 = 1.585 moles

Step 3: Calculate volume

For 1 mol at STP we have 22.4 L

For 1.585 moles we have 22.4 * 1.585 = 35.504 L

Option B is correct

Which of the following represents the MASS of 215 L of dinitrogen monoxide at STP? Hint...find the # of moles of this gas and then use the molar mass to go from moles to mass!

Step 1: Data given

Volume = 215 L

STP = 1 atm and 273 K

Molar mass of N2O = 44.01 g/mol

Step 2: Calulate moles N2O

22.4L = 1 mol at STP

215 L = 215 / 22.4 = 9.60 moles

Step 3: Calculate mass N2O

Mass N2O = moles N2O * molar mass N2O

Mass N2O = 9.60 moles* 44.01 g/mol

Mass N2O = 422.446 g

Option B is correct

You might be interested in

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

3 years ago

Most lakes have rivers flowing out of them, carrying water to the ocean. However, some lakes, including Great Salt Lake and the

Kaylis [27]

Since the only way of water flow to these lakes or bodies of water is through evaporation, I would expect an increase in unknown substances in the composition of the lakes due to the amount of contamination that globalization produces and affects terribly the surroundings when these unknown substances travel through evaporation as the outlet of these bodies of water. Therefore I think continuous contamination is what to expect after many more years of inflow and evaporation.

4 0

2 years ago

Consider the reaction 2 NO + O2 → 2 NO2

Sonbull [250]

Answer:

(a) Rate at which $NO_2$ is formed is 0.050 M/s

(b) Rate at which $O_2$ is consumed is 0.0250 M/s.

Explanation:

The given reaction is:-

$2NO+O_2\rightarrow 2NO_2$

The expression for rate can be written as:-

$-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}$

Given that:- $\frac{d[NO]}{dt}=-0.050\ M/s$ (Negative sign shows consumption)

$-\frac{1}{2}\frac{d[NO]}{dt}=\frac{1}{2}\frac{d[NO_2]}{dt}$

$-\frac{d[NO]}{dt}=\frac{d[NO_2]}{dt}$

$-(-0.050\ M/s)=\frac{d[NO_2]}{dt}$

$\frac{d[NO_2]}{dt}=0.050\ M/s$

(a) Rate at which $NO_2$ is formed is 0.050 M/s

$-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[O_2]}{dt}$

$-\frac{1}{2}\times -0.050\ M/s=-\frac{d[O_2]}{dt}$

$\frac{d[O_2]}{dt}=0.0250\ M/s$

(b) Rate at which $O_2$ is consumed is 0.0250 M/s.

5 0

3 years ago

In which kind of community do people mostly work in groups that generate ideas and provide services?

slega [8]

A.it is Postindustrial

8 0

3 years ago

Read 2 more answers

What's the principle of Atomic emission spectroscopy?

slega [8]

Answer:

Explanation:

The theory or working principle of Atomic Emission Spectroscopy involves the examination of the wavelengths of photons discharged by atoms and molecules as they transit from a high energy state to a low energy state. A characteristic set of wavelengths is emitted by each element or substance which depends on its electronic structure.

4 0

2 years ago