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timama [110]
3 years ago
5

Imagine you are celebrating a birthday in December and need balloons. The temperature in the

Chemistry
1 answer:
natka813 [3]3 years ago
8 0
So I think that this is very cool elsewhere think that it’s also a little bit cool so you’re selling birthday OK the pressure is definitely 25 and the 2 L so basically the answer is five so everything is great that’s the answer
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Can swords shatter from cold temperatures
sp2606 [1]
Depends on how the sword is made, what materials are used and temperature used but yes they can shatter.

When molecules cool down they stop vibrating and moving as much and so they "shrink" and the metal of the sword becomes brittle. sometimes they shrink at different phases which cause tension in the sword if this tension is strong enough it can cause the metallic bonds to break causing the sword to shatter.

hope that helps 
6 0
3 years ago
Use the ball and stick models above. If you had a gram of water and a gram of
Sladkaya [172]

Answer:

When the weather is nice, many people begin to work on their yards and homes. For many projects, sand is needed as a foundation for a walk or to add to other materials. You could order up twenty million grains of sand and have people really stare at you. You could order by the pound, but that takes a lot of time weighing out. The best bet is to order by the yard, meaning a cubic yard. The loader can easily scoop up what you need and put it directly in your truck.

Avogadro’s Number

It certainly is easy to count bananas or to count elephants (as long as you stay out of their way). However, you would be counting grains of sugar from your sugar canister for a long, long time. Atoms and molecules are extremely small – far, far smaller than grains of sugar. Counting atoms or molecules is not only unwise, it is absolutely impossible. One drop of water contains about 10 22 molecules of water. If you counted 10 molecules every second for 50 years without stopping you would have counted only 1.6 × 10 10 molecules. Put another way, at that counting rate, it would take you over 30 trillion years to count the water molecules in one tiny drop.

Explanation:

6 0
2 years ago
Why does the moon and the earth have the same mass
Evgesh-ka [11]

Answer:

Moon rocks contain few volatile substances (e.g. water), which implies extra baking of the lunar surface relative to that of Earth. The relative abundance of oxygen isotopes on Earth and on the Moon are identical, which suggests that the Earth and Moon formed at the same distance from the Sun.

Explanation:

3 0
2 years ago
Read 2 more answers
Please help!!!!
Tomtit [17]

Answer 1) Option D : Discrete


Explanation : As in the example the scientist studies about the effects of growing human populations on the biodiversity which is found in a region, and where each region was selected had a different population density of humans from 1 to 10 million per 10 square miles. Then in each region the number of different species that can be found was recorded. So here the numerical data was collected for different regions. So, we can conclude it as discrete because when the variable takes on a countable number of values it is called as discrete.


Answer 2) Option D : The entertainment industry


Explanation : When people had enough time and money with them there was no need of creating the communications industry as it was a need not a luxurious thing. With the creation of Internet industry it is somewhat irrelevant. They had enough of time and money so building a labor industry seems to be a lame choice. So, the option of entertainment industry suits the best.

6 0
3 years ago
Read 2 more answers
Butane (C4 H10(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , Hf = –393.5 kJ/mol ) and water (H2 O,
WITCHER [35]

Answer: Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

Explanation:

The chemical equation for the combustion of butane follows:

2C_4H_{10}(g)+4O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(8\times \Delta H^o_f_{CO_2(g)})+(10\times \Delta H^o_f_{H_2O(g)})]-[(1\times \Delta H^o_f_{C_4H_{10}(g)})+(4\times \Delta H^o_f_{O_2(g)})]

We are given:

\Delta H^o_f_{(C_4H_{10}(g))}=-125.6kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.82kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.6)+(4\times 0)]\\\\\Delta H^o_{rxn}=-5315kJ

Enthalpy of combustion (per mole) of C_4H_{10} (g) is -2657.5 kJ

6 0
3 years ago
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