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Effectus [21]
3 years ago
8

Write a Scientific question for how temperature change affects gas inside a container

Chemistry
1 answer:
natima [27]3 years ago
7 0
By scientific question, it really just means create a normal question. The example would be:
How will gas be affected if the temperature in an enclosed container is to be changed?

There are mannnyyy ways how to write a question for this, you just have to make sure you write what you are changing and how you are changing it, oh and make it sound smart haha.
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Write balanced equations for each of the processes, choosing from the following substances as reactants: BaCl2 O2 H2SO4 HNO3 C2H
marishachu [46]

Answer:

BaCl₂ and H₂SO₄

Explanation:

The reaction is the precipitation of BaSO₄ from solution.

This means that our reactants must contain the Ba ion and SO₄²⁻ ion.  The reactants that meet this criteria are the; BaCl₂ aand H₂SO₄

The reaction is given as;

BaCl₂ + H₂SO₄ --> BaSO₄ + 2HCl

8 0
4 years ago
If the temperature of the helium balloon were increased from 30°C to 35°C and the volume of the balloon only expanded from 0.47L
Cerrena [4.2K]

Answer:r u from ridge

Explanation:

Mr Wilson trippin

4 0
3 years ago
Write an equation that shows the formation of a copper (I) ion from a neutral copper atom
kumpel [21]
Cu ----> Cu(+) + 1 e-
5 0
3 years ago
Read 2 more answers
Preparation of Standard Buffer for Calibration of a pH Meter The glass electrode used in commercial pH meters gives an electrica
lana66690 [7]

Answer:

  • Mass of NaH₂PO₄·H₂O = 8.542 g
  • Mass of Na₂HPO₄ = 5.410 g

Explanation:

Keeping in mind the equilibrium:

H₂PO₄⁻ ↔ HPO₄⁻² + H⁺

We use the Henderson-Hasselbalch equation (H-H):

pH = pka + log\frac{[A^{-}]}{[HA]}

For this problem [A⁻] = [HPO₄⁻²] and [HA] = [H₂PO₄⁻]

From literature we know that pka = 7.21, from the problem we know that pH=7.00 and that

[HPO₄⁻²] + [H₂PO₄⁻] = 0.100 M

From this equation we can <u>express [H₂PO₄⁻] in terms of [HPO₄⁻²]</u>:

[H₂PO₄⁻] = 0.100 M - [HPO₄⁻²]

And then replace [H₂PO₄⁻] in the H-H equation, <u>in order to calculate [HPO₄⁻²]</u>:

7.00=7.21+log\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]} \\-0.21=log\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]}\\10^{-0.21} =\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]}\\0.616*(0.100M-[HPO4^{-2}])=[HPO4^{-2}]\\0.0616 M = 1.616*[HPO4^{-2}]\\0.03812 M =[HPO4^{-2}]

With the value of  [H₂PO₄⁻],<u> we calculate [HPO₄⁻²]</u>:

[HPO₄⁻²] + 0.0381 M = 0.100 M

[HPO₄⁻²] = 0.0619 M

Finally, using the concentrations, the volume, and the molecular weights; we can calculate the weight of each substance:

  • Mass of NaH₂PO₄·H₂O = 0.0619 M * 1 L * 138 g/mol = 8.542 g
  • Mass of Na₂HPO₄ = 0.0381 M * 1 L * 142 g/mol = 5.410 g

8 0
4 years ago
How to remember all your answers for a test?
Troyanec [42]

rewrite them and say them out loud or in your head repeatedly

8 0
4 years ago
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