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6 days ago

14

Emma wants to go swimming the first day the pool opens in May, but she is worried it will be too cold since summer only just sta

rted. She only likes to swim in the water if the pool is at least 82°F. The thermometer in the pool reads 24°C. Is the pool warm enough for Emma to swim in? Show all work to support your answer.

1 answer:

VMariaS [2.6K]6 days ago

6 0

Answer:

the temperature of the water is insufficient for her

Explanation:

Her desired temperature = 82°F

Water thermometer reading = 24°C

Convert the desired temperature to °C

°C=5/9 (82-32)

°C = 27.8 °C

Thus, the temperature of the water is not warm enough for her

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Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in

eduard [2520]

Clarification:

The pertinent information is outlined as follows.

m = 10.0 kg = 10,000 g (since 1 kg = 1000 g)

Starting temperature of block 1, $T_{1} = 100^{o}C$ = (100 + 273) K = 373 K

Starting temperature of block 2, $T_{2} = 0^{o}C$ = (0 + 273) K = 273 K

Therefore, the heat lost by block 1 equals the heat received by block 2

$mC \Delta T = mC \times \Delta T$

$10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C$

$T_{f} - 100^{o}C = 0^{o}C - T_{f}$

$2T_{f} = 100^{o}C$

$T_{f} = 50^{o}C$

It's important to convert the temperature into Kelvin as (50 + 273) K = 323 K.

Additionally, the relationship between enthalpy and temperature change is as follows.

$\Delta H = mC \Delta T$

= $10000 g \times 0.385 J/K g \times 323 K$

= 1243550 J

or, = 1243.5 kJ

Next, determine the entropy change for block 1 as follows.

$\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{373}$

= $10000 g \times 0.385 J/K g \times -0.143$

= -554.12 J/K

Now, the entropy change for block 2 is as follows.

$\Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{273}$

= $10000 g \times 0.385 J/K g \times 0.168$

= 647.49 J/K

Thus, the total entropy is the sum of the entropy changes of both blocks.

= -554.12 J/K + 647.49 J/K $\Delta S_{total} = \Delta S_{1} + \Delta S_{2}$

= 93.37 J/K

In conclusion, for this reaction, the outcome is 1243.5 kJ and $\Delta S_{total}$ is 93.37 J/K.

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11 days ago

(a) The original value of the reaction quotient, Qc, for the reaction of H2(g) and I2(g) to form HI(g) (before any reactions tak

KiRa [2726]

Response:

Here's my calculation

Clarification:

Assume the starting concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.

We need to determine the initial concentration of HI.

1. We will need a chemical equation with concentrations, so let's compile all the information in one location.

H₂ + I₂ ⇌ 2HI

I/mol·L⁻¹: 0.30 0.15 x

2. Calculate the concentration of HI

$Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}$

3. Plot the initial values

The graph below visualizes the initial concentrations as plotted on the vertical axis.

7 0

1 month ago

At STP, the element oxygen can exist as either O2 or O3 gas molecules. These two forms of the element have(1) the same chemical

Anarel [2605]

O2 and O3 represent different forms of Oxygen; therefore, they exhibit (4) distinct chemical and physical properties..

5 0

28 days ago

Read 2 more answers

Which of the following pairs lists a substance that can neutralize Ca(OH)2 and the salt that would be produced from the reaction

Anarel [2605]

Answer:

HCl and CaCl2

Explanation:

Calcium hydroxide acts as a base, thus requiring an acid for neutralization.

HCl functions as an acid, which can neutralize calcium hydroxide when they react together.

The resulting salt will be calcium chloride (CaCl2).

The interaction between calcium hydroxide and HCl yields Ca(OH)Cl and CaCl2.

Here, Ca(OH)Cl exists as a solution and calcium chloride serves as a base.

I hope this response assists you.

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25 days ago

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First-order reaction that results in the destruction of a pollutant has a rate constant of 0.l/day. (a) how many days will it ta

Alekssandra [2719]

The rate equation for a first order reaction can be expressed as follows:

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}$

In this context, k represents the reaction's rate constant, t denotes the time the reaction takes, $A_{0}$ is the initial concentration, and $A_{t}$ is the concentration at time t.

The rate constant of the reaction is $0.1 day^{-1}$ .

(a) If we start with an initial concentration of 100, when 90% of the substance is eliminated, the remaining quantity at time t will be 100-90=10. By substituting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days$

The time required to destroy 90% of the substance amounts to 23.03 days.

(b) If the initial concentration is set at 100, when 99% is destroyed, the present amount at time t will be 100-99=1. By substituting the input values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days$

This results in a duration of 46.06 days required to eradicate 99% of the chemical.

(c) Should the initial concentration be set at 100, with 99.9% of the chemical removed, the remaining quantity at time t will be 100-99.9=0.1. Substituting the values yields

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days$

Thus, the time needed to eliminate 99.9% of the chemical is calculated as 69.09 days.

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12 days ago