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alexandr1967 [171]
4 years ago
13

helium-filled balloon at sea level has a volume of 2.1 L at 0.998 atm and 36°C. If it is released and rises to an elevation at w

hich the pressure is 0.900 atm and the temperature is 28°C, what will be the new volume of the balloo
Chemistry
1 answer:
elena55 [62]4 years ago
8 0

the new volume of the balloon would be 2.3 L

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Which of the following systems has potential energy only?
DIA [1.3K]

Answer:

a bird sitting on a branch

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3 years ago
Copper sulfate can crystalize as an anhydrate (no water in the crystal) or as three different hydrates (varying amounts of water
Maurinko [17]

Answer:

Two factors that might have a affect of which copper sulphate mineral will occur at a given location  is:

A. Copper sulphate high solubility in water

B. Also it binds nicely with the sediments  or the crystal.

Explanation:

As it is mentioned here that copper sulphate can be crystallized as an anhydrate which means that their  is no waterin those crystals or can be as of those three different hydrates whose crystal structure varies with the amount of water present in it.

The four forms are also given of the copper sulphate are:

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So, the two factors that might give an affect which type of copper sulphate  mineral willoccur at a given location is:

A. The copper sulphate high solubility in water.

B. It binds extremely nicely with the sediments or say to the crystal. It is  also regulated by plants.

8 0
3 years ago
Please help
Colt1911 [192]
A i belive is the correct answer

3 0
3 years ago
What is the mole-to-mole ratio between N2 and H2?
spin [16.1K]
Remember, look at the coefficients in the balanced equation! Here are three mole ratios:
1 mole<span> N2 / </span>3 moles<span> H2.</span>
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</span>


5 0
3 years ago
12. What is the frequency of a photon with an energy of 3.03 x 10-19 J?
bazaltina [42]

Answer:

u=4.57x10^5GHz

Explanation:

Hello.

In this case, given the formula:

E=h*u

Whereas E is the energy, h the Planck's constant and u the frequency of the photon. Thus, solving for it, we obtain:

u=\frac{E}{h}=\frac{3.03x10^{-19}J}{6.63x10^{-34}J*s}\\  \\u=4.57x10^{14}s^{-1}

Or also:

u=4.57x10^{14}Hz*\frac{1GHz}{1x10^9Hz}\\ \\u=4.57x10^5GHz

Best regards.

5 0
3 years ago
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