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Anastaziya [24]
4 years ago
5

In Antoine Lavoisiers classic experiment, Mercuric oxide is heated in a sealed container. The solid red powder is change into tw

o products:Silver liquid mercury and oxygen gas.
If Lavoisier. Heated 100 g of powder mercuric oxide to produce 93 g of liquid mercury, how much oxygen would be released?

A) 7 g
B) 16 g
C) 32 g
D) 93 g
Chemistry
1 answer:
soldi70 [24.7K]4 years ago
7 0
Wha wha wa wa wa I'm sad wha!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! by the way the answer is B. 16 grams because the 93 g of murquiry wouldn't fit wha wha wha!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
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What would be the new pressure if 250. сm of gas at 25.0°C and 730. mmHg pressure
IrinaVladis [17]

Answer:

2340mmHg

Explanation:

P1V1/n1T1=P2V2/n2T2

n1=n2 T1=25+271=298k

T2=300+273=573k

730×250/298=P2×150/573

P2=2340mmHg

5 0
3 years ago
Which elements have 50protons
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Answer:

Tin (Sn) as in the periodic table the element tin has 50 protons

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What mass of H2SO, is required to react completely with 50.0 grams of ammonia (NH3)?​
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The answer is 10.7 grams
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3 years ago
What happens if a solid changes the state of substance to liquid?
kati45 [8]
It would had to have melted in order to change the properties from a solid into a liquid. It also depends on the type of solid wheither a chemical or a ice cube. If it was a ice cube if would dilutte the liquid solution and changes the taste ot texture (feel) if it was a chemical the properties could change the liquid into a solid only depending on th type of chemical. EX..Liquid Nitrogen and a piece of raw chiken the outcome after putting the chicken in the nitrogen would be frozen. Because the chemical Nitrogen freezes the particles which makes the chicken frozen.
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95. Using the standard enthalpy of formation data in Appendix G, calculate the bond energy of the carbon-sulfur double bond in C
prohojiy [21]

Answer:

+523 kJ.

Explanation:

The following data will be used to calculate the average C-S bond energy in CS2(l).

S(s) ---> S(g)

ΔH = 223 kJ/mol

C(s) ---> C(g)

ΔH = 715 kJ/mol

Enthalpy of formation of CS2(l)

ΔH = 88 kJ/mol

CS2(l) ---> CS2(g)

ΔH = 27 kJ/mol

CS2(g) --> C(g) + 2S(g)

So we must construct it stepwise.

1: C(s) ---> C(g) ΔH = 715 kJ

2: 2S(s) ---> 2S(g) ΔH = 446 kJ

adding 1 + 2 = 3

ΔH = 715 + 446

= 1161 kJ

3: C(s) + 2S(s) --> C(g) + 2S(g) ΔH = 1161 kJ

4: C(s) + 2S(s) --> CS2(l) ΔH = 88 kJ

adding (reversed 3) from 4 = 5

ΔH = -1161 + 88

= -1073 kJ

5: C(g) + 2S(g) --> CS2(l) ΔH = -1073 kJ

6: CS2(l) ---> CS2(g) ΔH = 27 kJ

adding 5 + 6 = 7

ΔH = -1073 + 27

= -1046 kJ

7. C(g) + 2S(g) --> CS2(g) ΔH = -1046 kJ

Reverse and divide by 2 for C-S bond enthalpy

= -(-1046)/2

= +523 kJ.

8 0
3 years ago
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