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2 years ago

In the reaction between the strong acid HCI and the strong base NaOH which of the ions listed below is a spectator ion?

Chemistry

2 answers:

qwelly [4]2 years ago

3 0

Answer:

Na+

Explanation:

The equation would be:

HCl (aq) + NaOH (aq) --> HOH (l) + NaCl (aq)

The equation is already balanced and the NaCl will disassociate in Na+ and Cl- and HCl will disassociate into H+ and Cl- and NaOH will disassociate into Na+ and OH-. Na+ is on both sides of the equation and stays the same, so Na+ will be the spectator ion.

Paladinen [302]2 years ago

3 0

Answer:

Na+

Explanation:

Spectator ions are ions that remain unchanged in a chemical reaction i.e. they exists in the same form in the reactant and the product side of the equation. Since these ions do not actually participate in the chemical reaction, they are termed as spectator ions.

The reaction between HCl and NaOH is a neutralization reaction yielding salt NaCl and water H2O.

HCl and NaOH are strong acids and bases respectively which completely dissociate into ions. NaCl is also an ionic compound which exists as Na+ and Cl-. The chemical reaction is:

$HCl + NaOH \rightarrow NaCl + H2O$

The total ionic equation is:

$H^{+} + Cl^{-} + Na^{+} + OH^{-} \rightarrow Na^{+} + Cl^{-} + H2O$

Based on the above equation: Na+ and Cl- ions remain the same on the reactant and product side. Hence, they are the spectator ions.

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A 100 gram glass container contains 200 grams of water and 50.0 grams of ice all at 0°c. a 200 gram piece of lead at 100°c is ad

ASHA 777 [7]

$0 \; \textdegree{\text{C}}$

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

$E(\text{fusion}) = m(\text{ice}) \cdot L_{f}(\text{water}) = 66.74 \; \text{kJ}$ and
$m(\text{water, final}) = m(\text{water, initial}) + m(\text{ice, initial}) = 0.250 \; \text{kg}$

Let the final temperature of the system be $t \; \textdegree{\text{C}}$ . Thus $\Delta T (\text{water}) = \Delta T (\text{beaker}) = t(\text{initial}) - t_{0} = t \; \textdegree{\text{C}}$

$Q(\text{water}) &= &c(\text{water}) \cdot m(\text{water, final}) \cdot \Delta T (\text{water})= 1.047 \cdot t\; \text{kJ}$ (converted to kilojoules)
$Q(\text{container}) &= &c(\text{glass}) \cdot m(\text{container}) \cdot \Delta T (\text{container})= 0.0837 \cdot t \; \text{kJ}$
$Q(\text{lead}) &= &c(\text{lead}) \cdot m(\text{lead}) \cdot \Delta T (\text{lead})= 0.0255 \cdot (100 - t)\; \text{kJ}$

The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable $t$ .

$E(\text{absorbed} ) = E(\text{released})$

$E(\text{absorbed} ) = E(\text{fushion}) + Q(\text{water}) + Q(\text{container})$
$E(\text{released}) = Q(\text{lead})$

Confirm the uniformity of units, equate the two expressions and solve for $t$ :

$66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)$

$t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}}$ which goes against the initial assumption. Implying that the final temperature does <em>not</em> go above the melting point of water- i.e., $t \le 0 \; \textdegree{\text{C}}$ . However, there's no way for the temperature of the system to go below $0 \; \textdegree{\text{C}}$ ; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at $0 \; \textdegree{\text{C}}$ ; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

3 0

3 years ago

What is the pH of a solution with an [H+] of (a) 5.4 x 10-10, (b) 4.3 x 10-5, (c) 5.4 x 10-7?

IgorLugansk [536]

Answer:

a. 9.2

b. 4.4

c. 6.3

Explanation:

In order to calculate the pH of each solution, we will use the definition of pH.

pH = -log [H⁺]

(a) [H⁺] = 5.4 × 10⁻¹⁰ M

pH = -log [H⁺] = -log 5.4 × 10⁻¹⁰ = 9.2

Since pH > 7, the solution is basic.

(b) [H⁺] = 4.3 × 10⁻⁵ M

pH = -log [H⁺] = -log 4.3 × 10⁻⁵ = 4.4

Since pH < 7, the solution is acid.

(c) [H⁺] = 5.4 × 10⁻⁷ M

pH = -log [H⁺] = -log 5.4 × 10⁻⁷ = 6.3

Since pH < 7, the solution is acid.

3 0

2 years ago

Compounds X and Y are both C6H13Cl compounds formed in the radical chlorination of 3-methylpentane. Both X and Y undergo base-pr

zubka84 [21]

Answer:

Y is a 3-chloro-3-methylpentane.

The structure is shown in the figure attached.

Explanation:

The radical chlorination of 3-methylpentane can lead to a tertiary substituted carbon (Y) and to a secondary one (X).

The E2 elimination mechanism, as shown in the figure, will happen with a simulyaneous attack from the base and elimination of the chlorine. This means that primary and secondary substracts undergo the E2 mechanism faster than tertiary substracts.