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2 years ago

I need help on #6 please

Chemistry

1 answer:

kobusy [5.1K]2 years ago

5 0

6. a) Team 3 (b/c their numbers were closer apart then the rest)
b)No they were not (can happen you might be precises but not accurate)
c)Team four, (on trial 4)

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The specific heat of copper metal is 0. 385 J/(g °C). How much energy must be added to a 35. 0-gram sample of copper to change t

Rus_ich [418]

The amount of heat required for changing the temperature of copper has been 606 J. Thus, option B is correct.

Specific heat has been defined as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius.

The heat required to raise the temperature has been expressed as:

$\rm Heat=mass\;\times\;specific\;heat\;\times\;Change\;in\;temperature$

<h3>Computation for the heat energy required</h3>

The given specific heat of copper has been $\rm 0.385\;J/g^\circ C$

The mass of copper has been, $\rm 35\;g$

The initial temperature of copper has been, $\rm 20^\circ C$

The final temperature of copper has been, $\rm 65^\circ C$

The change in temperature has been, $\Delta T$

$\Delta T=\text{Final\;temperature-Initial\;temperature}\\\Delta T =65^\circ \text C-20^\circ \text C\\\Delta T=45^\circ \text C$

Substituting the values for the heat required as:

$\rm Heat=35\;g\;\times\;0.385\;J/g^\circ C\;\times\;45^\circ C\\Heat=606\;J$

The amount of heat required for changing the temperature of copper has been 606 J. Thus, option B is correct.

Learn more about specific heat, here:

brainly.com/question/2094845

7 0

2 years ago

What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250

Blababa [14]

The buffer solution target has a pH value smaller than that of pKw (i.e., pH < 7.) The solution is therefore acidic. It contains significantly more protons $\text{H}^{+}$ than hydroxide ions $\text{OH}^{-}$ . The equilibrium equation shall thus contain protons rather than a combination of water and hydroxide ions as the reacting species.

Assuming that $x \; \text{L}$ of the 0.307 $\text{mol} \cdot \text{dm}^{-3}$ sodium hydroxide solution was added to the acetic acid. Based on previous reasoning, $x$ is sufficiently small that acetic acid was in excess, and no hydroxide ion has yet been produced in the solution. The solution would thus contain $0.2000 \times 0.425 - 0.307 \; x = 0.085 - 0.307 \; x$ moles of acetic acid and $0.307 \; x$ moles of acetate ions.

Let $\text{HAc}$ denotes an acetic acid molecule and $\text{Ac}^{-}$ denotes an acetate ion. The RICE table below resembles the hydrolysis equilibrium going on within the buffer solution.

$\begin{array}{lccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\end{array}$

The buffer shall have a pH of 4.250, meaning that it shall have an equilibrium proton concentration of $10^{4.250}\; \text{mol}\cdot \text{dm}^{-3}$ . There were no proton in the buffer solution before the hydrolysis of acetic acid. Therefore the table shall have an increase of $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ in proton concentration in the third row. Atoms conserve. Thus the concentration increase of protons by $10^{-4.250}\;\text{mol}\cdot \text{dm}^{-3}$ would correspond to a decrease in acetic acid concentration and an increase in acetate ion concentration by the same amount. That is:

$\begin{array}{lcccccc}\text{R} & \text{HAc} & \leftrightharpoons & \text{H}^{+} & + & \text{Ac}^{-}\\\text{I} & 0.085 - 0.307 \; x& & 0 & & 0.307 \; x\\\text{C} & - 10^{-4.250} & & +10^{-4.250} & & +10^{-4.250} \\\text{E} & 0.085 - 10^{-4.250} - 0.307 \; x& & 10^{-4.250} & & 10^{-4.250} + 0.307 \; x\end{array}$

By definition:

$\text{K}_{a} = [\text{H}^{+}] \cdot [\text{Ac}^{-}] / [\text{HAc}]\\\phantom{\text{K}_{a}} = 10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x)$

The question states that

$\text{K}_{a} = 1.75 \times 10^{-5}$

such that

$10^{-4.250} \times (10^{-4.250} + 0.307 \; x) / (0.085 - 10^{-4.250} - 0.307 \; x) = 1.75 \times 10^{-5}\\6.16 \times 10^{-5} \; x = 1.48 \times 10^{-6}\\x = 0.0241$

Thus it takes $0.0241 \; \text{L}$ of sodium hydroxide to produce this buffer solution.

6 0

3 years ago

Please help me answser this, its a photo<br><br>o element<br>o subscribe<br>o coefficient

Oliga [24]

Answer:

3 atoms of (C)carbon, 5 atoms of (H)hydrogen and 2 atoms of (O)oxygen

Explanation:

i don't know what you mean by subscribe

and i don't know what a coefficient is

4 0

3 years ago

How many moles are present in 4.7 liters of argon gas as STP

algol [13]

Answer:

0.21mol Ar (g)

Explanation:

To convert from litres to moles at STP we must divide the amount of litres by 22.4.

4.7 / 22.4 = 0.21mol Ar (g)

5 0

2 years ago

Read 2 more answers

Select the keyword or phrase that will best complete each sentence. A Claisen reaction is a nucleophilic ____________ in which a

Juliette [100K]

Answer:

Substitution.

Explanation:

Claisen reaction was first published in 1887 by a prominent German chemist known as Rainer Ludwig Claisen.

A Claisen reaction is a nucleophilic substitution in which an enolate is the nucleophile. It's typically a reaction of two molecules of an ester to form a β-keto ester, in the presence of an alkoxide base. Thus, a Claisen reaction is simply a characteristic condensation reaction of esters through a nucleophilic carbonyl substitution with an enolate such as a ketone enolate or an ester enolate.

Furthermore, a Claisen reaction results in the formation of a carbon-carbon bond in the presence of a strong base to yield a β-keto ester.

5 0

2 years ago