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2 years ago

Part F

1 answer:

7 0

Answer: did you ever find the answer? I need part E, F, and G if you found them please!!!!! I’m begging, my school year ends in 30 minutes and I gotta get this submitted!!!!!!

Explanation:

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What is the mass of 8.12 × 10^23 molecules of CO2 gas? (Atomic mass of carbon = 12.011 u; oxygen = 15.999 u.)

amid [387]

Answer:

$m=59.3gCO_2$

Explanation:

Hello,

In this case, the first step is to compute the molar mass of carbon dioxide as shown below, considering it has one carbon atom and two oxygen atoms:

$M=12.011g/mol+2*15.999g/mol\\\\M=44.009g/mol$

It is important to notice it is the mass in one mole of such compound. Afterwards, we need to use the Avogadro's number to compute the how many moles are in the given molecules of carbon dioxide as shown below:

$mol=8.12x10^{23}molec*\frac{1mol}{6.022x10^{23}molec} =1.35mol$

Finally, the mass by using the molar mass:

$m=1.35mol*\frac{44.009g}{1mol} \\\\m=59.3gCO_2$

Best regards.

4 0

2 years ago

How do muscles work in pairs to make parts of the body move?

schepotkina [342]

Answer:

Muscles work in pairs to move a bone.

Skeletal muscles only PULL in one direction. For this reason they always come in pairs. When one muscle in a pair contracts, to bend a joint for example, its counterpart then contracts and pulls in the opposite direction to straighten the joint out again.

4 0

2 years ago

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Calculate the molarity of a nitric acid solution if 38 ml of the solution is neutralized by 16 ml of 0.25 M barium hydroxide sol

RUDIKE [14]

Answer:

Explanation:

8 0

2 years ago

A gas with 4.0 atmospheres of pressure has a temperature of 27°C.

Gala2k [10]

Answer:

7.462

Explanation:

Well, every time that the tempurature is increased, the atmspheric pressure is increased by 0.574%. This would then mean that you would have 0.574 times

13. That would then equal 7.462. I hope this helps.

6 0

2 years ago

At a festival, spherical balloons with a radius of 280. Cm are to be inflated with hot air and released. The air at the festival

evablogger [386]

Answer:

8608.18 balloons

Explanation:

Hello! Let's solve this!

Data needed:

Enthalpy of propane formation: 103.85kJ / mol

Specific heat capacity of air: 1.009J · g ° C

Density of air at 100 ° C: 0.946kg / m3

Density of propane at 100 ° C: 1.440kg / m3

First we will calculate the propane heat (C3H8)

3000g * (1mol / 44g) * (103.85kJ / mol) * (1000J / 1kJ) = 7.08068 * 10 ^ 6 J

Then we can calculate the mass of the air with the heat formula

Q = mc delta T

m = Q / c delta T = (7.08068 * 10 ^ 6 J) / (1.009J / kg ° C * (100-25) ° C) =

m = 93566.96kg

We now calculate the volume of a balloon.

V = 4/3 * pi * r ^ 3 = 4/3 * 3.14 * 1.4m ^ 3 = 11.49m ^ 3

Now we calculate the mass of the balloon

mg = 0.946kg / m3 * 11.49m ^ 3 = 10.87kg

The amount of balloons is

93566.96kg / 10.87kg = 8608.18 balloons

5 0

3 years ago

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