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astra-53 [7]
3 years ago
8

3. How can you decrease the pressure of a gas in a container without changing the volume of the gas?

Chemistry
1 answer:
inna [77]3 years ago
8 0

Answer:

reduce the temperature of the gas

Explanation:

when you reduce the temperature of the gas the pressure will decrease

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Phytoplankton are microscopic photosynthesizing organisms that typically live in the upper layers of the ocean. What trophic lev
alex41 [277]

Answer:

Primary producers

Explanation:

Phytoplankton belong to the <u><em>primary producers</em></u> trophic level. Beings of the other levels gain energy from phytoplankton. Phytoplankton are also capable of transforming inorganic carbon into protoplasm.

4 0
4 years ago
A bowl of trail mix and a bowl of soup are examples of what type of matter? 9 (a) solution (b) heterogeneous mixture (c) element
AleksandrR [38]
The answer should be B.
6 0
3 years ago
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HELP!!! i will give brainliest!!
ICE Princess25 [194]

Answer:

Explanation:

fluorine have gained one electron that is why the sign is -1. they both have different number of protons. They have different neutron numbers. F have 10 and O have 8.

hope this helps :)

4 0
3 years ago
Uranium-232 has a half-life of 68.8 years. After 344.0 years, how much uranium-232 will remain from a 100.0-g sample?
amm1812

Answer:  3.13 g

Explanation:

Radioactive decay follows first order kinetics.

Half-life of uranium-232 = 68.8 years

\lambda =\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{68.8}= 0.010072674 year^{-1}

N=N_o\times e^{-\lambda t}

N = amount left after time t

N_0 = initial amount

\lambda = rate constant

t= time

N_0 = 100 g, t= 344 years, \lambda=0.010072674 years^{-1}

N=100\times e^{- 0.010072674 years^{-1}\times 344 years}

N=3.13g


8 0
3 years ago
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The 1995 Nobel Prize in chemistry was shared by Paul Crutzen, F. Sherwood Rowland, and Mario Molina for their work concerning th
Nikolay [14]

Answer:

\Delta H_{rxn3}=-162.5 kJ/mol

Explanation:

The reaction we need to calculate:

O_3 (g) + Cl (g) \longrightarrow ClO (g) + O_2 (g)

1) O_3 (g) + ClO (g) \longrightarrow Cl (g) +2 O_2 (g)

\Delta H_{rxn}=-122.8 kJ/mol

We need the ClO in the products side, so we use the inverse of this reaction:

Cl (g) +2 O_2 (g) \longrightarrow O_3 (g) + ClO (g)

\Delta H_{rxn1}=122.8 kJ/mol

2) 2 O_3 (g) \longrightarrow 3 O_2 (g)

\Delta H_{rxn2}=-285.3 kJ/mol

Now we need to combine this two:

Cl (g) +2 O_2 (g) + 2 O_3 (g)\longrightarrow O_3 (g) + ClO (g) + 3 O_2 (g)

Cl (g) + O_3 (g)\longrightarrow ClO (g) + O_2 (g)

The enthalpy of reaction:

\Delta H_{rxn3}=\Delta H_{rxn1}+ \Delta H_{rxn2}=122.8 kJ/mol-285.3 kJ/mol

\Delta H_{rxn3}=-162.5 kJ/mol

4 0
3 years ago
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