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Fantom [35]
3 years ago
11

The high concentration of carbon dioxide (CO2) in the atmosphere of Mars inhibits infrared radiation from escaping and keeps the

surface extremely hot.A. True/ VenusB. False/ Venus.
Chemistry
1 answer:
Zarrin [17]3 years ago
3 0

Answer:

B

Explanation:

Mars atmosphere contains mainly carbon iv oxide, but the greenhouse effect as subdued as there is so little CO2 overall.

Venus as a planet contains 96.5% of carbon iv oxide. it doesn't contain water which can trap the CO2 as we have on earth where the oceans are able to trap the CO2 present and subdue the greenhouse effect. This inability to trap CO2 in Venus prevents infrared rays from escaping and with the fact the Venus is closer to the sun than earth and mars, its surface it extremely hot.

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How many neutrons would be in an atom of Selenium if the mass number for this particular isotope is Se-80?
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Which one of the following descriptions relating to nuclear reactions is correct?
monitta

Answer: option D - The total number of nucleons changes.

Explanation:

Nuclear Reaction is best described as a process such as the fission of an atomic nucleus, or the fusion of one or more atomic nuclei and / or subatomic particles in which the NUMBER of PROTONS and / or NEUTRONS in a nucleus CHANGES; the reaction products may contain a different element or a different isotope of the same element.

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4 0
3 years ago
Which of the following statements describes
kozerog [31]

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Energy is released as heat during the reaction.

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3 years ago
What is the composition, in atom percent, of an alloy that contains a) 45.5 lbm of silver, b) 83.7 lbm of gold, and c) 6.3 lbm o
lyudmila [28]

Answer:

\% atAg=44.6\%\\\% atAu=44.9\%\\\% atCu=10.5\%

Explanation:

Hello,

In this case, for computing the atom percent, one must obtain the number of atoms of silver, gold and copper as shown below:

atomsAg=45.5lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{107.87gAg}*\frac{6.022x10^{23}atomsAg}{1molAg}=1.15x10^{26}atomsAg\\atomsAu=83.7lbm*\frac{453.59g}{1lbm}*\frac{1molAu}{196.97gAu}*\frac{6.022x10^{23}atomsAu}{1molAu}=1.16x10^{26}atomsAu\\atomsCu=6.3lbm*\frac{453.59g}{1lbm}*\frac{1molAg}{63.55gCu}*\frac{6.022x10^{23}atomsCu}{1molCu}=2.71x10^{25}atomsCu

Thus, the atom percent turns out:

\% atAg=\frac{1.15x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.6\%\\\% atAu=\frac{1.16x10^{26}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =44.9\%\\\% atCu=\frac{2.71x10^{25}}{1.15x10^{26}+1.16x10^{26}+2.71x10^{25}}*100\% =10.5\%

Best regards.

4 0
3 years ago
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