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astra-53 [7]
3 years ago
8

Terry heats two different masses of substance A to different temperatures, and measures the amounts of both carbon dioxide and w

ater produced.
Which statement is true about this experiment?

This experiment is uncontrolled because two different masses of substance A are used.

This experiment is well controlled because only the mass of substance A is varied.

This experiment is uncontrolled because two gaseous products are measured.

This experiment is well controlled because the masses of substance A and the products are varied.
Chemistry
2 answers:
Naddika [18.5K]3 years ago
5 0

Answer:

This experiment is uncontrolled because two different masses of substance A are used.

Explanation:

A controlled experiment is a structured experiment aimed at testing a particular observation or observations. The setup of a controlled experiment helps to determine the reason why a particular observation occurs and what must have led to it.

In the experiment highlighted above, different masses of a substance were used, they were heated to different temperatures. The set up does not show any correlation between the masses of substances heated and the temperatures. It is even difficult to try to predict the hypothesis for this kind of experimental set up. All the variables in play can best be assumed to be independent of one another.

Svetach [21]3 years ago
5 0

Answer:

This experiment is uncontrolled because two different masses of substance A are used.

Explanation:

done the topic

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Why do atoms fond bonds​
Setler [38]
They bond because they want to make their outer electron shells more stable

Hope this helps

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3 0
3 years ago
Two common materials we encounter in our daily lives are radiator coolant in automobiles and dry ice. The coolant in an automobi
Yuki888 [10]

Answer:

See explanation

Explanation:

Using the formula

°C = (F-32) × 5/9

Where;

°C = temperature in degrees centigrade

F= temperature in Fahrenheit

F= (9/5 ×°C) +32

F= (9/5 × 110) + 32

F= 230°F

To convert -78°C to Kelvin

-78°C + 273 = 195 K

3 0
3 years ago
What must be the molarity of an aqueous solution of trimethylamine, (ch3)3n, if it has a ph = 11.20? (ch3)3n+h2o⇌(ch3)3nh++oh−kb
Stolb23 [73]

0.040 mol / dm³. (2 sig. fig.)

<h3>Explanation</h3>

(\text{CH}_3)_3\text{N} in this question acts as a weak base. As seen in the equation in the question, (\text{CH}_3)_3\text{N} produces \text{OH}^{-} rather than \text{H}^{+} when it dissolves in water. The concentration of \text{OH}^{-} will likely be more useful than that of \text{H}^{+} for the calculations here.

Finding the value of [\text{OH}^{-}] from pH:

Assume that \text{pK}_w = 14,

\begin{array}{ll}\text{pOH} = \text{pK}_w - \text{pH} \\ \phantom{\text{pOH}} = 14 - 11.20 &\text{True only under room temperature where }\text{pK}_w = 14 \\\phantom{\text{pOH}}= 2.80\end{array}.

[\text{OH}^{-}] =10^{-\text{pOH}} =10^{-2.80} = 1.59\;\text{mol}\cdot\text{dm}^{-3}.

Solve for [(\text{CH}_3)_3\text{N}]_\text{initial}:

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}

Note that water isn't part of this expression.

The value of Kb is quite small. The change in (\text{CH}_3)_3\text{N} is nearly negligible once it dissolves. In other words,

[(\text{CH}_3)_3\text{N}]_\text{initial} = [(\text{CH}_3)_3\text{N}]_\text{final}.

Also, for each mole of \text{OH}^{-} produced, one mole of (\text{CH}_3)_3\text{NH}^{+} was also produced. The solution started with a small amount of either species. As a result,

[(\text{CH}_3)_3\text{NH}^{+}] = [\text{OH}^{-}] = 10^{-2.80} = 1.58\times 10^{-3}\;\text{mol}\cdot\text{dm}^{-3}.

\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\textbf{initial}} = \text{K}_b = 1.58\times 10^{-3},

[(\text{CH}_3)_3\text{N}]_\textbf{initial} =\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{\text{K}_b},

[(\text{CH}_3)_3\text{N}]_\text{initial} =\dfrac{(1.58\times10^{-3})^{2}}{6.3\times10^{-5}} = 0.040\;\text{mol}\cdot\text{dm}^{-3}.

8 0
3 years ago
How does acid affect limestone?
Andrej [43]

Answer:

It reacts by fizzing and wearing away/dissolving the rock.

Explanation:

7 0
3 years ago
Which in bigger 43 mg or 5 g
AleksandrR [38]
5 g is bigger than 43 mg.

I hope this is right, I apologize if I'm wrong.
6 0
3 years ago
Read 2 more answers
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