Metals exhibit

To what category of elements does an element belong if it is a poor conductor of electricity

Yuri [45]

I was wondering the same thing my dude

3 0

3 years ago

100 POINTS ANSWER HURRY

prisoha [69]

C. Not all mixtures have solutes and solvents

7 0

4 years ago

Read 2 more answers

A student balanced the chemical equation Mg + O2 →MgO by writing Mg + O2 → MgO2. Was the equation balanced correctly? Explain yo

Amanda [17]

Explanation:

Charges on both magnesium and oxygen is 2. Though opposite in sign, they have equal charges so, both of them will be cancelled by each other.

As a result, formula of magnesium oxide is MgO and not $MgO_{2}$ .

The student write the equation as $Mg + O_2 \rightarrow MgO_2$ , it is not correct.

Therefore, given equation will be balanced as follows.

$2Mg + O_{2} \rightarrow 2MgO$

Since, number of atoms on both reactant and product side are equal. Hence, this equation is completely balanced.

7 0

3 years ago

A 62.6-gram piece of heated limestone is placed into 75.0 grams of water at 23.1°C. The limestone and the water come to a final

VMariaS [17]

M₁ = mass of water = 75 g
T₁ = initial temperature of water = 23.1 °C
c₁ = specific heat of water = 4.186 J/g°C

m₂ = mass of limestone = 62.6 g
T₂ = initial temperature of limestone = ?
c₂ = specific heat of limestone = 0.921 J/g°C

T = equilibrium temperature = 51.9 °C
using conservation of heat
Heat lost by limestone = heat gained by water
m₂c₂(T₂ - T) = m₁c₁(T - T₁)
inserting the values
(62.6) (0.921) (T₂ - 51.9) = (75) (4.186) (51.9 - 23.1)
T₂ = 208.73 °C
in three significant figures
T₂ = 209 °C

7 0

4 years ago

Two solutions namely, 500 ml of 0.50 m hcl and 500 ml of 0.50 m naoh at the same temperature of 21.6 are mixed in a constant-pre

weeeeeb [17]

24.6 ℃

<h3>Explanation</h3>

Hydrochloric acid and sodium hydroxide reacts by the following equation:

$\text{HCl} \; (aq) + \text{NaOH} \; (aq) \to \text{NaCl} \; (aq) + \text{H}_2\text{O} \; (aq)$

which is equivalent to

$\text{H}^{+} \; (aq) + \text{OH}^{-} \; (aq) \to \text{H}_2\text{O}\; (l)$

The question states that the second equation has an enthalpy, or "heat", of neutralization of $-56.2 \; \text{kJ}$ . Thus the combination of every mole of hydrogen ions and hydroxide ions in solution would produce $56.2 \; \text{kJ}$ or $56.2 \times 10^{3}\; \text{J}$ of energy.

500 milliliter of a 0.50 mol per liter "M" solution contains 0.25 moles of the solute. There are thus 0.25 moles of hydrogen ions and hydroxide ions in the two 0.500 milliliter solutions, respectively. They would combine to release $0.25 \times 56.2 \times 10^{3} = 1.405 \times 10^{4} \; \text{J}$ of energy.

Both the solution and the calorimeter absorb energy released in this neutralization reaction. Their temperature change is dependent on the heat capacity <em>C</em> of the two objects, combined.

The question has given the heat capacity of the calorimeter directly.

The heat capacity (the one without mass in the unit) of water is to be calculated from its mass and <em>specific</em> heat.

The calorimeter contains 1.00 liters or $1.00 \times 10^{3} \; \text{ml}$ of the 1.0 gram per milliliter solution. Accordingly, it would have a mass of $1.00 \times 10^{3} \; \text{g}$ .

The solution has a specific heat of $4.184 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}$ . The solution thus have a heat capacity of $4.184 \times 1.00 \times 10^{3} = 4.184 \times 10^{3} \; \text{J} \cdot\text{K}^{-1}$ . Note that one degree Kelvins K is equivalent to one degree celsius ℃ in temperature change measurements.

The calorimeter-solution system thus has a heat capacity of $4.634 \times 10^{3} \; \text{J} \cdot \text{K}^{-1}$ , meaning that its temperature would rise by 1 degree celsius on the absorption of 4.634 × 10³ joules of energy. $1.405 \times 10^{4} \; \text{J}$ are available from the reaction. Thus, the temperature of the system shall have risen by 3.03 degrees celsius to 24.6 degrees celsius by the end of the reaction.

4 0

3 years ago