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koban [17]
2 years ago
5

1. At STP, how many liters of oxygen are required to react completely with 3.6 liters of hydrogen to form water? 2H2(g)+O2(g) -&

gt; 2H2O(g)
a. 1.8 L
b. 3.6 L
c. 2.0 L
d. 2.4 L

2. Which conversion factor do you use first to calculate the number of grams of CO2 produced by the reaction of 50.6 g of CH4 with O2? The equation for the complete combustion of methane is: CH4(g) +2O2(g) -> CO2 + 2H2O(l)

a. 1 mol CH4/ 16.0g CH4
b. 2 mol O2/1 mol CO2
c. 16.0g CH4/1 mol CO4
d. 44.0g CO2/2 mol CO2

3. Which of the following statements is true about the following reaction? 3NaHCO3(aq) + C6H8)7(aq) -> 3CO2(g) + 3H2O(s) + Na3C6H5O7(aq)

a. 22.4 L of CO2(g) are produced for every liter of C6H8O7(aq) reacted.
b. 1 mole of water is produced for every mole of carbon dioxide produced.
c. 6.02 x 10^23 molecules of Na3C6H5O7(aq) are produced for every mole of NaHCO3(aq) used.
d. 54 g of water are produced for every mole of NaHC)3(aq) produced.
Chemistry
1 answer:
ElenaW [278]2 years ago
6 0
1. is a
2.is a
3.is b
hope this helps
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Please help
inna [77]

From the calculation, the standard free energy of the system is -359kJ.

<h3>What is the standard free-energy?</h3>

The  standard free-energy is the energy present in the system. We have to first obtain the cell potential using the formula;

Ereduction - E oxidation = 0.96 V - 0.34 V = 0.62 V

Using the formula;

ΔG = -nFEcell

ΔG =-(6 * 96500 * 0.62)

ΔG =-359kJ

Learn more about free energy:brainly.com/question/15319033

#SPJ1

3 0
1 year ago
By what factor will the kinetic energy change if the speed of the baseball is decreased to 54.8 mi/h? Express your answer as an
Serggg [28]

Answer:

The Kinetic Energy is approximately 3 times decreased

Explanation:

A baseball weighs 5.13 oz.  

a)What is the kinetic energy, in joules, of this baseball when it is thrown by a major league pitcher at 95.o mi/h?  

b) By what factor will the kinetic energy change if the speed of the baseball is decreased to 54.8 mi/h? Express your answer as an integer.

Kinetic Energy (KE)=0.5×mass×velocity ^ 2

Kinetic Energy (KE)=0.5×mass × velocity ^ 2

Joules = kg×m^2/s^2

1 mile = 1609.344 meters

1 hour = 3600 sec

1 Oz = 28.34952 g = 0.02834952 kg

a) KE=0.5×m×v^2

=0.5×(5.13 oz × 0.02834952 kg/1 ounce)×(95 miles/h × 1609.344 m/1 mile × 1 hr/3600 s)^2

=130.761 kg×m^2/s^2 = 130.761 Joules

b) KE=0.5×m×v^2

=0.5×(5.13 oz × 0.02834952 kg/1 ounce)×(54.8 miles/h × 1609.344 m/1 mile × 1 hr/3600 s)^2

=43.51028 kg×m^2/s^2 = 43.51028 Joules

= 130.761 / 43.51028 = 3.00528,  

As such the Kinetic Energy is approximately 3 times decreased

4 0
3 years ago
Fill in the Blank
Arisa [49]

Answer: Quantitative data

Explanation:

3 0
3 years ago
Explain the ways that carbon dioxide is added to the atmosphere. How is it removed? Does most carbon enter the atmosphere as car
Licemer1 [7]

Answer:

Carbon dioxide is added to the atmosphere by human activities. When hydrocarbon fuels (i.e. wood, coal, natural gas, gasoline, and oil) are burned, carbon dioxide is released. During combustion or burning, carbon from fossil fuels combine with oxygen in the air to form carbon dioxide and water vapor.Carbon moves from fossil fuels to the atmosphere when fuels are burned. When humans burn fossil fuels to power factories, power plants, cars and trucks, most of the carbon quickly enters the atmosphere as carbon dioxide gas. Each year, five and a half billion tons of carbon is released by burning fossil fuels.Carbon dioxide causes about 20 percent of Earth's greenhouse effect; water vapor accounts for about 50 percent; and clouds account for 25 percent.Likewise, when carbon dioxide concentrations rise, air temperatures go up, and more water vapor evaporates into the atmosphere—which then amplifies greenhouse heating

5 0
2 years ago
What is the concentration of a solution with a volume of 660L that contains 33.4g of AlCO3?
lana [24]

Answer:

The concentration of the solution is 5.8168 × 10^{-4} mol.dm^{-3}

Explanation:

Here, we want to calculate the concentration of the solution.

The unit of this is mol/dm^3

So the first thing to do here is to calculate the number of moles of the solute present, which is the number of moles of AlCO3

The number of moles = mass/molar mass

molar mass of AlCO3 = 27 + 12 + 3(16)  = 27 + 12 + 48 = 87g/mol

Number of moles = 33.4/87 = 0.384 moles

This 0.384 moles is present in 660 L

x moles will be present in 1 dm^3

Recall 1 dm^3 = 1L

x * 660 = 0.384 * 1

x = 0.384/660 = 0.00058168 = 5.8168 * 10^-4 mol/dm^3

7 0
2 years ago
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