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andrey2020
13 days ago
8

At what temperature would the volume of a gas be 0.550 L if it had a volume of 0.432 L at –20.0 o C?

Chemistry
2 answers:
castortr0y [927]13 days ago
8 0
To find the temperature at which the volume of the gas would be 0.550 L, given that it is 0.432 L at -20.0 °C, apply Charles’s Law.

The formula is v1/T1 = v2/T2
Known values:
V1 = 0.550 L
T1 = ?
T2 = -20°C + 273 = 253 K
V2 = 0.432 L

Rearranging for T1:
T1 = (V1 × T2) / V2

Calculating:
T1 = (0.55 L × 253) / 0.432 L = 322.11 K or 49.11°C
eduard [944]13 days ago
4 0

Answer: 321.6 K

Explanation: The gas behavior follows Charles' Law, which states that at constant pressure and mass, volume is directly proportional to temperature.

V\propto T (assuming constant pressure and moles)

\frac{V_1}{T_1}=\frac{V_2}{T_2}

Where:

V_1 = initial volume = 0.432 L

V_2 = final volume = 0.550 L

T_1 = initial temperature = -20^oC=273-20=253K

T_2 = final temperature = unknown

Substituting known values into the formula yields final temperature:

\frac{0.432L}{253K}=\frac{0.550L}{T_2}

T_2=321.6K

Thus, the final temperature is 321.6 K.

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Calculate the molarity of 48.0 mL of 6.00 M H2SO4 diluted to 0.250 L
lorasvet [960]

Answer:

The molality is 1.15 m.

Molality is calculated by dividing the number of moles of solute by the kilograms of solvent, which in this case is water.

Calculate moles of H₂SO₄ from molarity:

C = n/V → n = C × V = 6.00 mol/L × 0.048 L = 0.288 moles

Mass of solvent (water) based on density:

m = ρ × V = 1.00 kg/L × 0.250 L = 0.250 kg

Therefore, molality is:

m = moles/solvent mass = 0.288 moles / 0.250 kg = 1.15 m

4 0
14 days ago
Read 2 more answers
(a) calculate the %ic of the interatomic bond for the intermetallic compound tial3. (b) on the basis of this result, what type o
Tems11 [854]

Answer :

The percentage ionic character (%IC) equals 10%, indicating the bond is mostly covalent with slight polarity.

Percent Ionic Character:

This reflects the fraction of ionic nature within a polar covalent bond. The formula for %IC (% ionic character) is:

Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^X^a^-^X^b^) * 100

Here, Xa is the electronegativity of atom A and Xb is that of atom B.

Given: The compound is TiAl₃.

Electronegativity of Ti = 2.0

Electronegativity of Al = 1.6 (as shown in the provided image)

Substitute these values into the formula:

Percent Ionic character = 1 - e^-^0^.^2^5 ^*^(^2^.^0^-^1^.^6^) * 100

Percent Ionic character = 1 - e^(^-^0^.^2^5 ^*^0^.^4^) * 100

Percent Ionic character = 1 - e^(^-^0^.^1^) * 100

The value of e⁻¹ equals 0.90.

Therefore, percent ionic character = (1 - 0.90) × 100

Percent Ionic Character = 10%

Because the % IC is only 10%, which is relatively low, the bond is classified as covalent with minimal polarity.

8 0
16 days ago
A 25.0 g sample of an alloy was heated to 100.0 oC and dropped into a beaker containing 90 grams of water at 25.32 oC. The tempe
KiRa [976]

Response:

The specific heat of the alloy C_{a} = 0.37 \frac{KJ}{Kg K}

Clarification:

Weight of the alloy m_{a} = 25 gm

Initial temperature T_{a} = 100°c = 373 K

Weight of the water m_{w} = 90 gm

Initial temperature of water T_{w} = 25.32 °c = 298.32 K

Final temperature T_{f} = 27.18 °c = 300.18 K

Using the energy balance equation,

Heat released by the alloy = Heat absorbed by the water

m_{a} C_{a} [[T_{a} - T_{f}] = m_{w} C_w (T_{f} -T_{w} )

25 × C_{a} × ( 373 - 300.18 ) = 90 × 4.2 (300.18 - 298.32)

C_{a} = 0.37 \frac{KJ}{Kg K}

This gives us the specific heat of the alloy.

4 0
6 days ago
What features of this model will help Armando answer the question?
lions [1003]

Answer:

The adjustable legs along with the sand table.

Note: The question is incomplete. The full question is presented below.

Using Models to Address Questions Regarding Systems

Armando’s class was examining images of rivers shaped by flowing water. Most rivers appeared wide and shallow, except for one, which was narrow and deep. The students theorized that this river's narrowness and depth are due to:

  • the steepness of the hill from which the water descends, or
  • the diminutive size of the sand grains the water flows through.

To explore the answer to the question of why this river is so narrow and deep, Armando created the model outlined below.

Explanation:

The model constructed by Armando will facilitate addressing the question due to specific features:

1. Adjustable leg - as one theory proposed by the class suggests that the steep hill affecting the water's path could be the reason for the river's dimensions, the adjustable legs are designed to be raised or lowered to alter the slope, allowing testing of this theory.

2. Sand table - this acts as the streambed. By modifying the size of the sand grains, students can examine the second hypothesis that smaller sand grains contribute to the river's narrowness and depth.

The outcomes of their experimentation will lead them to a conclusion.

5 0
5 days ago
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So, could you use easily use chromic acid to visualize spots when you are monitoring a reaction to make a ketone out of an alcoh
VMariaS [1037]

While the original inquiry is incomplete, the comprehensive question is:

Many chemicals can illustrate spots on a TLC plate that have been processed and dried. The permanganate used in the video creates yellow spots against a purplish background, taking advantage of the oxidizing capability of basic permanganate (MnO4), which outperforms chromic acid as an oxidizing agent. Chromic acid can also be employed to visualize spots, resulting in a green hue on the yellow background, indicating oxidation. So, can chromic acid be conveniently used to visualize spots when tracking a reaction converting an alcohol into a ketone? What observations are anticipated if one attempts this? Furthermore, if a small amount of alcohol is included in a solvent mixture for eluting your TLC plate, why must the plate be fully dried before visualizing the spots with an oxidizing agent like permanganate or chromic acid?

Answer:

Typically, using chromic acid to visualize spots during the conversion of alcohol to ketone is not feasible. The alcohol (substrate) will convert into its respective ketone due to the presence of chromic acid, causing the spots for the product and the reactant to align horizontally. This alignment complicates differentiation between the spots, making chromic acid unsuitable for this purpose.

It's vital to ensure that the plate is completely dry before observing spots with an oxidizing agent, even if alcohol is present in the solvent mixture. Incomplete drying could lead to oxidation of the alcohol by the oxidizing agent, resulting in transformation to carboxylic acid or ketone, thereby creating a new spot.

6 0
13 days ago
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