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Whitepunk [10]
3 years ago
11

What happens when plaster of Paris is treated with co2

Chemistry
1 answer:
umka21 [38]3 years ago
5 0
) adding phosphoric acid, said phosphoric acid reacting with said calcium carbonate to release gaseous carbon dioxide bubbles, and said water providing water of hydration to said plaster of Paris to form gypsum that sets around said carbon dioxide bubbles thereby forming said foamed plaster.
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Please help me with this or I may get an F in chemistry :(
umka21 [38]

Answer:

Subconscious - psychic activity just below the level of awareness

Election - the act of electing someone

Explanation:

I hope this helped enough, good luck in chemistry!

8 0
3 years ago
C3H8+3O2 = 3CO2+4H2O what is the enthalpy combustion
dezoksy [38]

Answer:

\Delta H_{comb}=2043.85kJ/mol

Explanation:

Hello there!

In this case, according to the given chemical reaction, it possible for us to set up the expression for the calculation of the enthalpy change as shown below:

\Delta H_r=-\Delta H_{comb}=3\Delta _fH_{CO_2}+4\Delta _fH_{H_2O}-\Delta _fH_{C_3H_8}-3\Delta _fH_{O_2}

Thus, given the values of the enthalpies of formation on the attached file, we obtain:-\Delta H_{comb}=3(-393.5kJ/mol)+4(-241.8kJ/mol)-(-103.85kJ/mol)-3(0kJ/mol)\\\\-\Delta H_{comb}=-2043.85kJ/mol\\\\\Delta H_{comb}=2043.85kJ/mol

Best regards!

8 0
3 years ago
By what method do we measure the distance to objects that are a few thousand light-years away?
Zigmanuir [339]

Answer:

i believe the answer to your question is parallax or parsecs. im sorry im not very specific in this!

Explanation:

8 0
3 years ago
Read 2 more answers
How does water heat earth
vovikov84 [41]
The answer is the 3rd one down I think
6 0
4 years ago
How many days does it take 16.Og of Gold-198 to decay to 1.0g? (each half-
forsale [732]

Answer:

10.8 days (3 sig.figs.)

Explanation:

All radioactive decay is 1st order decay defined by the expression A = A₀e^-kt

which is solved for time of decay (t) => t = ln(A/A₀) / -k

A = final weight = 1.0 gram

A₀ = initial weight = 16.0 grams

k = rate constant = 0.693/t(1/2) = 0.693/2.69 days = 0.258 days⁻¹

t = ln(1/16) / -0.258da⁻¹ = (-2.77/-0.258) days = 10.74646792 days (calculator)

≅ 10 days (1 sig. fig. based on given 1 gram mass)

4 0
3 years ago
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