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stepan [7]
4 years ago
6

Bromine, a liquid at room temperature, has a boiling point of 58°C and a melting point of –7.2°C. Bromine can be classified as a

Chemistry
2 answers:
daser333 [38]4 years ago
6 0
<span>Answer: option C) pure substance.
</span><span />

<span>Justification:
</span><span />

<span>A sharp melting point is a characteristic of pure substances.
</span><span />

<span>The pure substances are either elements or compounds: they have a unique chemical constitution (kind of elements, number of atoms of each element in the chemical formula, and same chemical bonds).
</span><span>
</span><span>
</span><span> Other products, this is mixtures (either homogeneus or heterogeneous), being composed of different elements and compounds in different variable ratios, exhibit a range as boiling point. For example the boiling point of the gasoline is about 19°C to 200°C.
</span>

Bromine is an element: you find it in the periodic table, with the atomic number 35, hence a pure substance.
<span /><span />
Mama L [17]4 years ago
5 0

On mine the answer is (D) but the answer would still be "pure substance."

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If a sample of sodium chloride with a mass of
alex41 [277]
Original molarity was 1.7 moles of NaCl

Final molarity was 0.36 moles of NaCl

Given Information:

Original (concentrated) solution: 25 g NaCl in a 250 mL solution, solve for molarity

Final (diluted) solution: More water is added to make the new total volume 1.2 liters, solve for the new molarity

1. Solve for the molarity of the original (concentrated) solution.

Molarity (M) = moles of solute (mol) / liters of solution (L)

Convert the given information to the appropriate units before plugging in and solving for molarity.

Molarity (M) = 0.43 mol NaCl solute / 0.250 L solution = 1.7 M NaCl (original solution)

2. Solve for the molarity of the final (diluted) solution.

Remember that the amount of solute remains constant in a dilution problem; it is just the total volume of the solution that changes due to the addition of solvent.

Molarity (M) = 0.43 mol NaCl solute / 1.2 L solution

Molarity (M) of the final solution = 0.36 M NaCl

I hope this helped:))
6 0
3 years ago
The ksp of copper(ii) carbonate, cuco3, is 1.4 × 10-10. calculate the molar solubility of this compound.
mafiozo [28]
As,
                              CuCO₃    ⇆  Cu²⁺  +  CO₃²⁻
So,
                             Kc  =  [Cu²⁺] [CO₃²⁻] / CuCO₃
Or,
                             Kc (CuCO₃)  =  [Cu²⁺] [CO₃²⁻]
Or,
                             Ksp  =  [Cu²⁺] [CO₃²⁻]
As,
Ksp  = 1.4 × 10⁻¹⁰
So,
                            1.4 × 10⁻¹⁰  =  [x] [x]
Or,
                             x²  =  1.4 × 10⁻¹⁰ 
Or,
                             x  =  1.18 × 10⁻⁵ mol/L
To cahnge ito g/L,
                             x  =  1.18 × 10⁻⁵ mol/L × 123.526 g/mol

                   
         x  =  1.45 × 10⁻³ g/L
3 0
3 years ago
A small piece of hot metal is placed in cooler water. The metal is left in the water
scoundrel [369]

Answer: The amount of energy lost by the metal is equal to the amount of energy gained by the water

Explanation:

5 0
2 years ago
What geometry does VSEPR predict for the central atom in NCl3
Archy [21]

Answer:

Tetrahedral electron geometry and trigonal pyramidal molecular geometry.

Explanation:

The Lewis structure is shown in Figure 1.

The central N atom has three bonding pairs and one lone pair, for <em>four electron groups</em>.

VSEPR theory predicts a tetrahedral electron geometry with bond angles of 109.5°.

We do not count the lone pair in determining the molecular shape.

The molecular geometry is trigonal pyramidal (see Figure 2).

5 0
3 years ago
4
STALIN [3.7K]

Answer: The space occupied by the gas at 400 torr and 25^{o}C is 250 mL.

Explanation:

Given: V_{1} = 250 mL,    P_{1} = 800 torr,     T_{1} = 50^{o}C

V_{2} = ?,         P_{2} = 400 torr,        T_{2} = 25^{o}C

Formula used is as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{800 torr \times 250 mL}{50^{o}C} = \frac{400 torr \times V_{2}}{25^{o}C}\\V_{2} = 250 mL

Thus, we can conclude that space occupied by the gas at 400 torr and 25^{o}C is 250 mL.

7 0
3 years ago
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