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andriy [413]
3 years ago
6

What kind of rocks might you expect to find on the Moon? Why do you think that?​

Chemistry
1 answer:
Vinil7 [7]3 years ago
3 0

Answer:

The Moon's surface is dominated by igneous rocks. The lunar highlands are formed of anorthosite, an igneous rock predominantly of calcium-rich plagioclase feldspar

Explanation:

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What is the de Broglie wavelength, in cm, of a 11.0-g hummingbird flying at 1.20 x 10^2 mph?
KonstantinChe [14]

Answer:1.123 x 10^-31cm

Explanation:

mass of humming bird=  11.0g

speed= 1.20x10^2mph

but I mile = 1.6m

1km=1000

I mile = 1.6x10^3m

1.20x10^2mph= 1.6x10^3m /1mile x at 1.20 x 10^2

=1.932 x10^5m

recall that  

1 hr= 60 min

1 min=60 secs, 1hr=3600s

Speed = distance/ time

=1.932 x10^5 / 3600= 5.366 x 10 ^1 m/s

m= a 11.0g= 11.0 x 10^-3kg

h=6.626*10^-34 (kg*m^2)/s

Wavelength = h/mu

= 6.626*10^-34/(11 x 10^-3 x 5.366x 10^1)

6.63x10^-34/ 590.26x 10 ^-3= 1.123 x10^-33m

but 1m = 100cm

1.123 x 10 ^-33 x 100 = 1.123 x 10^-31cm

de broglie wavelength of humming bird = 1.123 x 10 ^-31cm

5 0
3 years ago
A chemist wants to observe the following reaction:
stiv31 [10]

It take more energy to break the bonds of the reactants and less energy is given off when the product bonds are formed.

<h3>What is Energy?</h3>

Energy is defined as the ability to do work. Work is done in the breaking or formation of bonds.

The standard Enthalpy (ΔH) of water which was formed in the given reaction is negative.

ΔH= Δproduct - Δreactant

This means that the energy to break the bonds of the reactants is more.

Read more about Enthalpy here brainly.com/question/14291557

3 0
2 years ago
So its not A so its B C or D plz help
Rainbow [258]

Answer:

the answer is B

hope this helped!

3 0
3 years ago
For a particular reaction, ΔH=−111.4 kJ/mol and ΔS=−25.0 J/(mol·K).
PtichkaEL [24]
That’s the answer ^^^^
3 0
3 years ago
A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s
Lilit [14]

Answer:

pH = 1.32

Explanation:

                 H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount  to buffer solutions.  In the first  deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g  =  0.074 mol H₂M

54 mL x  1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

           

3 0
3 years ago
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