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3 years ago

Why did engineers need to design a sunshade for Mercury Messenger?

Chemistry

2 answers:

zloy xaker [14]3 years ago

8 0

The answer is so it didn't burn.

scoundrel [369]3 years ago

6 0

They designed one cause of the sun powerful radiant energy and thermal they needed to protect the satellite from being disintegrated when it gets close to the sun its also the closet plant to the sun MVEMJSUN is a mnemonic device I made up it means my very energetic mother just served us nachos or that's what I say and that is your answer bam

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A three-carbon alkene contains two double bonds. Predict its molecular formula? A. C3H4 B. C3H5 C. C3H6

mestny [16]

The correct answer is A. C3H4 is the three-carbon alkene contains two double bonds. Alkenes are a class of organic compounds with carbon-carbon double bond. The structure for C3H4 is:

H2C=C=CH2

7 0

4 years ago

8. How much heat will be released when

Eva8 [605]

Answer:

15 is it

Explanation:

3 0

3 years ago

Read 2 more answers

Which of the following nuclei would be the most stable?

ValentinkaMS [17]

Ten protons would be Neon, but the most stable? I'd have to say 12 neutrons in contrast to the 10 protons. Not too negative in balance.

6 0

4 years ago

2. Calculate the pl of the following amino acids(use their Pka values) a. Arginine b. Glutamic acid of water an c. Asparagine d.

Savatey [412]

<h2>♨ANSWER♥</h2>

pl (25*C)

Arginine -----> 10.76

Glutamic -----> 3.08

Asparagine -----> 5.43

Tyrosine -----> 5.63

☆...hope this helps...☆

_♡_mashi_♡_

5 0

2 years ago

A quantity of 0.0250 mol of a gas initially at 0.050 L and 19.0°C undergoes a constant-temperature expansion against a constant

KiRa [710]

Answer:

$V_2=2.995L\\\\W=248.5J$

Explanation:

Hello,

In this case, for us to compute the final volume we apply the Boyle's law that analyzes the pressure-volume temperature as an inversely proportional relationship:

$P_1V_1=P_2V_2$

So we solve for $V_2$ by firstly computing the initial pressure:

$P_1=\frac{nRT}{V_1}=\frac{0.025mol*0.082\frac{atm*L}{mol*K}*(19+273.15)K}{0.050L} =11.98atm$

$V_2=\frac{P_1V_1}{P_2}=\frac{11.98atm*0.050L}{0.200atm}\\ \\V_2=2.995L$

Finally, we can compute the work by using the following formula:

$W=nRTln(\frac{V_2}{V_1} )=0.025mol*8.314\frac{J}{mol*K}*(19.0+273.15)K*ln(\frac{2.995L}{0.050L}) \\\\W=248.5J$

Best regards.

4 0

4 years ago