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2 years ago

True or false. Air masses mix at fronts?

Chemistry

1 answer:

Elina [12.6K]2 years ago

6 0

The correct answer is False

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Finally, it is time to return to Earth and review how scientists here classify organisms:

lilavasa [31]

5. Eubacteria

6. Plantae

7. Animalia

8. Protist (technically not a kingdom)

9. Archaebacteria

10. Fungi

6 0

2 years ago

A 12.5 g sample of granite initially at 82.0 oc is immersed into 25.0 g of water that is initially at 22.0 oc. what is the final

Paraphin [41]

The amount of heat lost by granite is equal to the amount of heat gained by water. Therefore their change in enthalpies must be equal. The opposite in sign means that one is gaining while the other is losing

ΔH granite = - ΔH water

ΔH is the change in enthalpy experienced by a closed object as it undergoes change in energy. This is expressed mathematically as,

ΔH = m Cp (T2 – T1)

Given this information, we can say that:

12.5 g * 0.790 J / g ˚C * (T2 – 82 ˚C) = - 25.0 g * 4.18 J / g ˚C * (T2 – 22 ˚C)

9.875 (T2 – 82) = 104.5 (22 – T2)

9.875 T2 – 809.75 = 2299 – 104.5 T2

114.375 T2 = 3108.75

T2 = 27.18 ˚C

The temperature of 2 objects after reaching thermal equilibrium is 27.18 ˚C.

7 0

3 years ago

When sound molecules "stack up" in front of high-speed flights, what could

FrozenT [24]

pretty sure its B thank me later

4 0

3 years ago

On a hiking trip, you see a large tree root has grown into the cracks of a rock and caused it to break apart. This is an example

Vadim26 [7]

Hydrologic Weathering I think

3 0

3 years ago

g If 50.0 mL of a 0.75 M acetic acid solution is titrated with 1.0 M sodium hydroxide, what is the pH after 10.0 mL of NaOH have

V125BC [204]

Answer:

pH = 2.66

Explanation:

Acetic Acid + NaOH → Sodium Acetate + H₂O

First we calculate the number of moles of each reactant, using the given volumes and concentrations:

0.75 M Acetic acid * 50.0 mL = 37.5 mmol acetic acid

1.0 M NaOH * 10.0 mL = 10 mmol NaOH

We calculate how many acetic acid moles remain after the reaction:

37.5 mmol - 10 mmol = 27.5 mmol acetic acid

We now calculate the molar concentration of acetic acid after the reaction:

27.5 mmol / (50.0 mL + 10.0 mL) = 0.458 M

Then we calculate [H⁺], using the following formula for weak acid solutions:

[H⁺] = $\sqrt{C*Ka}=\sqrt{0.458M*1.76x10^{-5}}$

[H⁺] = 0.0028

Finally we calculate the pH:

pH = -log[H⁺]

pH = 2.66

8 0

2 years ago