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2 years ago

11

The Temperature of the earth gets__ as you get closer to the _ and __ as you move towards the_

1 answer:

bulgar [2K]2 years ago

4 0

The temperature of the earth gets HOTTER as you get closer to the CORE and COOLER as u move toward the SURFACE.

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Chemistry check please... Thank you!

I am Lyosha [343]

1) We apply the ideal gas equation:
PV = nRT
n = (21300 x 3/1000) / (8.314 x 323)
n = 0.024
Your answer is correct.

2) Total pressure = Partial pressure of Hydrogen + Partial pressure of water
134.7 = 122.4 + Pw
Pw = 12.3 kPa
Your answer is correct

3) The molar fraction, volume fraction and pressure fraction of gasses are the same thing.
Thus, percentage pressure of Oxygen = 10%
Pressure of Oxygen = 2.04 x 10⁴ x 0.1
= 2.04 x 10³ kPa
Your answer is correct

Well done!

3 0

2 years ago

Please help me with the questions?

weqwewe [10]

Answer:

1. Forecast

2. Cold front

3. Shelf Cloud

4. Barometer

5. Pressure

6 Prevailing Westerlies

7. Thermometer

8. Weather

6 0

2 years ago

You weigh two different objects on the same scale in the same location on Earth. The weights are different. What must be differe

timofeeve [1]

They can be differnt shape or one must weight more then the other one

3 0

2 years ago

Read 2 more answers

CzHs + __02-> __CO2+ _H20 *<br> I need help ASAP PLSSS

Alex777 [14]

Answer:

2,7,4,6

Explanation:

atoms at the left side should be equal to that of right for an equation to be balanced. so the answer is the final one

8 0

2 years ago

You are given 25.00 mL of an acetic acid solution of unknown concentration. You find it requires 35.75 mL of a 0.1950 M NaOH sol

oee [108]

Answer:

0.2788 M

1.674 %(m/V)

Explanation:

Step 1: Write the balanced equation

NaOH + CH₃COOH → CH₃COONa + H₂O

Step 2: Calculate the reacting moles of NaOH

$0.03575 L \times \frac{0.1950mol}{L} = 6.971 \times 10^{-3} mol$

Step 3: Calculate the reacting moles of CH₃COOH

The molar ratio of NaOH to CH₃COOH is 1:1.

$6.971 \times 10^{-3} molNaOH \times \frac{1molCH_3COOH}{1molNaOH} = 6.971 \times 10^{-3} molCH_3COOH$

Step 4: Calculate the molarity of the acetic acid solution

$M = \frac{6.971 \times 10^{-3} mol}{0.02500L} =0.2788 M$

Step 5: Calculate the mass of acetic acid

The molar mass of acetic acid is 60.05 g/mol.

$6.971 \times 10^{-3} mol \times \frac{60.05g}{mol} =0.4186 g$

Step 6: Calculate the percentage of acetic acid in the solution

$\frac{0.4186g}{25.00mL} \times 100\% = 1.674 \%(m/V)$

6 0

2 years ago

Read 2 more answers