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bazaltina [42]
2 years ago
6

Carbon monoxide (CO) gas reacts with oxygen (O2) gas to produce carbon dioxide (CO2) gas. If 1.00 L of carbon monoxide reacts wi

th excess oxygen at standard temperature and pressure, what volume of carbon dioxide is produced? What mass of carbon dioxide is produced?
Chemistry
2 answers:
rosijanka [135]2 years ago
6 0

Answer:

1.00 L

1.96 g

Explanation:

fredd [130]2 years ago
4 0
The balanced reaction would be:

2CO + O2 = 2CO2

We assume that the gases are ideal gas so that we use the relation that 1 mol of an ideal gas is equal to 22.4 L of the gas at STP. From that relation, we get the number of moles and we can convert it to other units. We do as follows:

1.0 L CO ( 1 mol / 22.4 L ) ( 2 mol CO2 / 2mol CO ) = 0.045 mol CO2 produced

0.045 mol CO2 ( 22.4 L / 1 mol ) = 1 L of CO2
0.045 mol CO2 ( 44.01 g / 1 mol ) = 1.98 g of CO2
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A piece of iron metal is heated to 155 degrees C and placed into a calorimeter that contains 50.0 mL of water at 18.7 degrees C.
Korvikt [17]

Answer:

D = 28.2g

Explanation:

Initial temperature of metal (T1) = 155°C

Initial Temperature of calorimeter (T2) = 18.7°C

Final temperature of solution (T3) = 26.4°C

Specific heat capacity of water (C2) = 4.184J/g°C

Specific heat capacity of metal (C1) = 0.444J/g°C

Volume of water = 50.0mL

Assuming no heat loss

Heat energy lost by metal = heat energy gain by water + calorimeter

Heat energy (Q) = MC∇T

M = mass

C = specific heat capacity

∇T = change in temperature

Mass of metal = M1

Mass of water = M2

Density = mass / volume

Mass = density * volume

Density of water = 1g/mL

Mass(M2) = 1 * 50

Mass = 50g

Heat loss by the metal = heat gain by water + calorimeter

M1C1(T1 - T3) = M2C2(T3 - T2)

M1 * 0.444 * (155 - 26.4) = 50 * 4.184 * (26.4 - 18.7)

0.444M1 * 128.6 = 209.2 * 7.7

57.0984M1 = 1610.84

M1 = 1610.84 / 57.0984

M1 = 28.21g

The mass of the metal is 28.21g

3 0
2 years ago
This stadium can hold 100,000, or 1 x 105, people. The number of atoms in a grain of iron is about 1 x 1018. Would you need 1 x 
nata0808 [166]
<h3>Answer:</h3>

1 x 10^13 stadiums

<h3>Explanation:</h3>

We are given that;

1 stadium holds = 1 × 10^5 people

Number of iron atoms is 1 × 10^18 atoms

Assuming the stadium would carry an equivalent number of atoms as people.

Then, 1 stadium will carry 1 × 10^5 atoms

Therefore,

To calculate the number of stadiums that can hold 1 × 10^18 atoms we divide the total number of atoms by the number of atoms per stadium.

Number of stadiums = Total number of atoms ÷ Number of atoms per stadium

                                  = 1 × 10^18 atoms ÷ 1 × 10^5 atoms/stadium

                                   = 1 × 10^13 Stadiums

Thus, 1 × 10^18 atoms would occupy 1 × 10^13 stadiums

7 0
1 year ago
What quantum numbers specify these subshells? 2S, 6P, and 3D. (The answer is n= and L=)
statuscvo [17]
N = 1
l = from 0 to (n-1)
ml = -1... + 1
ms = 1/2 or -1/2

eg = 2s
n = 2, m = 0, n = 0
s = 1/2, -1/2

hope this help
4 0
2 years ago
A new student is planning to use thin layer chromatography (TLC) for his research project. After setting up the apparatus the st
tamaranim1 [39]

Answer:

The open system evaporates the solvent in the solution

Explanation:

An open system is a system in which exchange of materials and energy can occur. If a TLC set up is left open, then the set up constitutes an open system.

During TLC, the sample is dotted on the plate and inserted into a suitable solvent. The solvent moves up the plate and achieves the required separation of the mixture.

Most of these solvents used used TLC are volatile organic compounds. Therefore, if the TLC set up is left open, the solvent will evaporate leading to poor results after running the TLC experiment.

7 0
2 years ago
A precipitate forms when mixing solutions of sodium fluoride (NaF) and lead II nitrate (Pb(NO3)2). Complete and balance the net
Margarita [4]

Answer:

Pb^2+(aq) + 2F-(aq) → PbF2(s)

Explanation:

Step 1: Data given

sodium fluoride = NaF

lead(II)nitrate Pb(NO3)2

Step 2: The unbalanced equation

NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + NaNO3(aq)

Step 3: Balancing the equation

NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + NaNO3(aq)

On the left side we have 2x NO3 (in Pb(NO3)2), on the right side we have 1x NO3 (in NaNO3). To balance the amount of NO3 we hvae to multiply NaNO3 on the right side by 2.

NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + 2NaNO3(aq)

On the left side we have 1x Na (in NaF), on the right side we have 2x Na (in 2NaNO3). To balance the amount of Na we have to multiply NaF on the left side by 2. Now the equation is balanced.

2NaF(aq) + Pb(NO3)2(aq) →  PbF2(s) + 2NaNO3(aq)

Step 4: Calculate net ionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2Na+(aq) + 2F-(aq) + Pb^2+(aq) + 2NO3-(aq) → PbF2(s) + 2Na+(aq) + NO3-(aq)

Pb^2+(aq) + 2F-(aq) → PbF2(s)

4 0
2 years ago
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