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Sergio039 [100]
3 years ago
8

A thermometer is placed in water in order to measure the water’s temperature. What would cause the liquid in the thermometer to

rise?
The molecules in the water move closer together.

The molecules in the thermometer’s liquid spread apart.

The kinetic energy of the water molecules decreases.

The kinetic energy of the thermometer’s liquid molecules decreases
Chemistry
2 answers:
borishaifa [10]3 years ago
3 0
I believe the correct answer from the choices listed above is the second option. It would be that the molecules in the thermometer’s liquid spread apart that caused <span>the liquid in the thermometer to rise. The heat expanded the liquid inside. Hope this answers the question.</span>
kaheart [24]3 years ago
3 0

Answer: Option (b) is the correct answer.

Explanation:

When we use a thermometer to measure the temperature of water then heat from the water will lead to represent an increase in the thermometer reading.

This is because heat or warmth from the water will get transferred to the thermometer as a result, the liquid in the thermometer will also get warm.

Therefore, this warmth will increase kinetic energy of the molecules of liquid in the thermometer. Hence, the liquid will spread.

Thus, we can conclude that the liquid in the thermometer to rise because  molecules in the thermometer’s liquid spread apart.

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They are all stable and have eight valence electrons
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3 years ago
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Which type of carbon fixation stores carbon dioxide in acid form? a. c3 b. c4 c. cam d. all of the above
Luba_88 [7]

The type of carbon fixation stores carbon dioxide in acid form is CAM i.e. crassulacean acid metabolism.

<h3>What are CAM?</h3>

CAM stands for crassulacean acid metabolism in this process photosynsthesis is occured at day time but the exchange of gases takes place at night itself only.

In this carbon fixation process, carbon dioxide is stored in the form of organic acid malic acid and losses carbon dioxide at the night time and by doing this it helps in the storage of water.

Hence option (C) is correct.

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8 0
2 years ago
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When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2
MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is \Delta \ T_f = T_{solvent}- T_{solution}

\Delta \ T_f = 2.9 ^0 \ C

Using the depression in freezing point, the molar depression constant of the solvent K_f = \dfrac{\Delta T_f}{m}

K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}

K_f = 4.82 ^0 C / m}

The freezing point of the solution \Delta T_f = T_{solvent} - T_{solution}

\Delta T_f = 7.3^ 0 \ C

The molality of the solution is:

= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}

Molar depression constant of solvent X, K_f = 4.82 ^0 \ C/m

Hence, using the elevation in boiling point;

the Vant'Hoff factor i = \dfrac{\Delta T_f}{k_f \times m}

i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}

\mathbf {i = 1.51 }

3 0
3 years ago
A 25.0-mL sample of 0.150 M hydrocyanic acid is titrated with a 0.150 M NaOH solution. The Ka of hydrocyanic acid is 4.9 × 10-10
lara [203]

Answer:

The pOH = 1.83

Explanation:

Step 1: Data given

volume of the sample = 25.0 mL

Molarity of hydrocyanic acid = 0.150 M

Molarity of NaOH = 0.150 M

Ka of hydrocyanic acid = 4.9 * 10^-10

Step 2: The balanced equation

HCN + NaOH → NaCN + H2O

Step 3: Calculate the number of moles hydrocyanic acid (HCN)

Moles HCN = molarity * volume

Moles HCN = 0.150 M * 0.0250 L

Moles HCN = 0.00375 moles

Step 3: Calculate moles NaOH

Moles NaOH = 0.150 M * 0.0305 L

Moles NaOH = 0.004575 moles

Step 4: Calculate the limiting reactant

0.00375 moles HCN will react with 0.004575 moles NaOH

HCN is the limiting reactant. It will completely be reacted. There will react 0.00375 moles NaOH. There will remain 0.004575 - 0.00375 = 0.000825 moles NaOH

Step 5: Calculate molarity of NaOH

Molarity NaOH = moles NaOH / volume

Molarity NaOH = 0.000825 moles / 0.0555 L

Molarity NaOH = 0.0149 M

Step 6: Calculate pOH

pOH = -log [OH-]

pOH = -log (0.0149)

pOH = 1.83

The pOH = 1.83

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