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iogann1982 [59]
3 years ago
12

You are working in your laboratory and decide to do some cleaning. You find a test tube with some brown substance congealed at t

he bottom. You take a piece of iron wool to scrub it out, but cannot remove it. You leave the iron wool in the test tube and set it aside to clean your beakers instead. You do not realize the test tube is next to the heater and after a few minutes, a brown gas starts smoking out of the test tube, followed by a reddish-brown gas. Upon observation, the iron wool is coated in a yellow-brown substance. You realize that the mystery substance was _________.
A. Argon (Ar)
B. Scandium (Sc)
C. Bromine (Br)
D. Iodide (I)
Chemistry
1 answer:
Anna [14]3 years ago
7 0

Answer:

  • <em>The mystery substance is</em> <u>C. Bromine (Br) </u>

Explanation:

<em>Argon (Ar) </em>is a noble gas. Whose freezing point is -189 °C (very low), thus it cannot be the frozen substance. Also, it is not reactive, thus is would have not reacted with iron. Hence, argon is not the mystery substance.

<em>Scandium (Sc) </em>is a metal from group 3 of the periodic table, thus is will not react with iron. Thus, scandium is not the mystery substance.

Both <em>bromine</em> and <em>iodine</em> are halogens (group 17 of the periodic table).

The freezing point of bromine is −7.2 °C, ​and the freezing point of iodine is 113.7 °C. Thus, both could be solids (frozen) in the lab.

The reactivity of the halogens decrease from top to bottom inside the group. Bromine is above iodine. Then bromine is more reactive than iodine.

Bromine is reactive enough to react with iron. Iodine is not reactive enough to react with iron.

You can find in the internet that bromine vapour over hot iron reacts  producing iron(III) bromide. Also, that bromine vapors are red-brown.

Therefore, <em>the mystery substance is bromine (Br).</em>

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Carbon-12 and carbon-13 are isotopes of carbon.
ohaa [14]

Answer:

Isotopes of an element share the same number of protons but have different numbers of neutrons. Let's use carbon as an example. There are three isotopes of carbon found in nature – carbon-12, carbon-13, and carbon-14. All three have six protons, but their neutron numbers - 6, 7, and 8, respectively - all differ.

Explanation:

4 0
3 years ago
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Solutions of aluminum nitrate and sodium sulfate react to form sodium nitrate and aluminum sulfate. Classify this reaction.
Temka [501]
C is the correct answer. Double displacement reaction, also called as metathesis, is a type of reaction wherein two compounds react to form new compounds. This case the cations and the anions of the reactants replace each other forming the new compounds.
4 0
3 years ago
Elements that have atoms with stable valence electron configurations in the ground state are found in
dsp73

Answer:

Group 18, also known as the Noble Gasses

Explanation:

Atoms strive for full stability by gaining or losing electrons to get 8 valence electrons in their valence shell, but Group 18 already has 8 electrons in their valence shell, and are therefore already stable in their ground state.

5 0
3 years ago
A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

5 0
3 years ago
The object is traveling at a velocity of m/s.
marishachu [46]

Answer:

= 3.75m {s}^{ - 1}  \\

Explanation:

velocity =  \frac{displacement}{time} \\  =  \frac{15m}{4s}   \\  = 3.75m {s}^{ - 1}

3 0
3 years ago
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