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Ber [7]
3 years ago
9

Mario uses a hot plate to heat a beaker of 50mL of water. He used a thermometer to measure the temperature of the water. The wat

er in the beaker began to boil when it reached the temperature of 100 C. If Mario completes the same experiment with 25mL of water, what would happen to the boiling point?
Chemistry
2 answers:
Alex777 [14]3 years ago
8 0

Answer:

The boiling point decreases as the volume decreases.

Explanation:

The Temperature - Volume law otherwise called as Charles law is applied, which says that the volume of the given gas at constant pressure is directly proportional to the temperature measured in Kelvin. As the volume increases, the temperature also increases, if the volume decreases, then the temperature also decreases.

As per the Charles law, here the volume is decreased from 50 ml to 25 ml so the boiling point also decreases.

Readme [11.4K]3 years ago
4 0

Answer:

nothing

Explanation:

as long as he's using pure water, there will be no change to the boling temperature. boiling point depends on the purity of the substance, not volume. so the temperature at which mario's water boils remains 100C.

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In the laboratory a student finds that it takes 44.0 Joules to increase the temperature of 10.6 grams of solid zinc from 24.9 to
BaLLatris [955]

Answer:

The specific heat of zinc is 0.361 J/g°C

Explanation:

<u>Step 1:</u> Data given

44.0 J needed

Mass of solid zinc = 10.6 grams

Initial temperature = 24.9 °C

Final temperature = 36.4 °C

<u>Step 2</u>: Calculate the specific heat of zinc

Q = m*c*ΔT

⇒ with Q = heat (in Joule) = 44.0 J

⇒ with m = the mass of the solid zinc = 10.6 grams

⇒ with c = the specific heat of the zinc = TO BE DETERMINED

⇒ with ΔT = The change in temperature = T2-T1 = 36.4 °C - 24.9 °C = 11.5 °C

44.0 J = 10.6 grams * c * 11.5°C

c = 44.0 J / (10.6g * 11.5 °C)

c = 0.361 J/g°C

The specific heat of zinc is 0.361 J/g°C

7 0
3 years ago
The alkali earth metal beryllium (Be) engages in a chemical reaction and loses all of its valence electrons.
denpristay [2]

The loss of electron from an results in the formation of cation represented by the positive charge on the element whereas gaining of electron results in the formation of anion represented by the negative charge on the element.

The alkali earth metal beryllium (Be) belongs to the second group of the periodic table. The ground state electronic configuration of Be is:1s^{2}2s^{2}

From the electronic configuration it is clear that it has 2 valence electrons in its valence shell (2s^{2}).

After losing all valence electrons that is 2 electrons from 2s orbital. The electronic configuration will be:

1s^{2}2s^{0}

Since, lose of electron is represented by positive charge on the element symbol. So, the beryllium will have +2 charge on its symbol as Be^{2+}.

Hence, beryllium will have 2+ charge on it after losing all its valence electrons in the chemical reaction.

6 0
3 years ago
Read 2 more answers
Please help make sure its correct thanks
Wewaii [24]

The empirical formula of metal iodide : CoI₃(Cobalt(III) Iodide)

<h3>Further explanation</h3>

13.02 g sample of Cobalt , then mol Co(MW=58.933 g/mol) :

\tt mol=\dfrac{mass}{Ar}=\\\\mol=\dfrac{13.02~g}{58.933}\\\\mol=0.221

Mass of metal iodide formed : 97.12 g, so mass of Iodine :

\tt =mass~metal~iodide-mass~Cobalt\\\\=97.12-13.02\\\\=84.1~g

Then mol iodine (MW=126.9045 g/mol) :

\tt \dfrac{84.1}{126.9045}=0.663

mol ratio of Cobalt and Iodine in the compound :

\tt 0.221\div 0.663=1\div 3

5 0
2 years ago
Which of the following statements are correct regarding the binding energy?
Andrews [41]
None of the above?
(Is there any statements?)
8 0
3 years ago
Match the condensed structural diagrams with the correct names
AleksAgata [21]

Answer:

Here's what I get  

Explanation:

1. Names

I.   CH₃-CH₂-COOH              = 49. propanoic acid

II.  CH₃-CH₂-OH                    = 46. ethanol

III. CH₃-COO-CH₂-CH₂-CH₃ = 47. propyl ethanoate

IV. H-O-CH₂-CH₂-CH₃         = 48. propan-1-ol

V.  H-COO-CH₃                    = 51. methyl methanoate

VI. CH₃-COOH                      = 50. ethanoic acid

2. Precursors

52. methyl propionate  ⇒ methanol + propanoic acid

53.   ethyl methanoate ⇒    ethanol + methanoic acid

6 0
3 years ago
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