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KonstantinChe
1 month ago
15

26.2 g piece of copper metal is heated from 21.5°C to 201.6°C. Calculate the amount of heat absorbed by the metal. The specific

heat of Cu is 0.385 /g "C.

Chemistry
1 answer:
lorasvet [2.7K]1 month ago
6 0

Response:1816.6 joules

Clarification:refer to the attached image

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Compound X contains only carbon, hydrogen, nitrogen, and chlorine. When 1.00 g of X is dissolved in water and allowed to react w
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Let's assume that the compound formula is as follows: Experiment 1: 1.00 g of the compound yields 1.95 g of AgCl. The molar mass of AgCl is 143.32 g/mol. Thus, the moles of AgCl for 1.95g are: The moles of Cl also equal 0.0136, considering that 1 mole of AgCl corresponds to 1 mole of Cl. Experiment 2: 1.00 g of the compound results in 0.900 g of CO2 and 0.735 g of H2O. The molar mass of CO2 is 44 g/mol, and for H2O, it's 18 g/mol. Therefore, the moles of C come to 0.0205 and the moles of H stand at 0.0816 (which is 2 times the moles of H2O). Now, from the provided details, it's derived that in 1.00 g of the compound, there are 0.0136 moles of Cl, 0.0205 moles of C, and 0.0816 moles of H. In terms of mass: Mass of Cl = 0.0136 * 35.5 = 0.4828 g. Mass of C = 0.0205 * 12 = 0.246 g. Mass of H = 0.0816 * 1 = 0.0816 g. Total mass = 0.4828 + 0.246 + 0.0816 + mass of N. Given that 1.00 = 0.8104 + Mass of N, it follows that Mass of N = 0.1896. Thus, upon dividing all moles by the smallest value, we find Cl = 0.0136 / 0.0135 = 1.0007; C = 0.0205 / 0.0135 = 1.52; H = 0.0816 / 0.0135 = 6.04; N = 0.0135 / 0.0135 = 1. Multiplying by 2 allows us to reach integer values: Cl = 2, C = 3, H = 12, N = 2.
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2 months ago
Hydrogen gas has a density of 0.090 g/L, and at normal pressure and -1.72 C one mole of it takes up 22.4 L. How would you calcul
Anarel [2989]

Answer:

n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}

Explanation:

Assuming all calculations occur at standard pressure and a temperature of -1.72°C :

n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}

Where

n is the number of moles of hydrogen

n is the mass of hydrogen

\rho is the density of hydrogen

6 0
3 months ago
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Thallium-207 decays exponentially with a half life of 4.5 minutes. if the initial amount of the isotope was 28 grams, how many g
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An exponential decay law is generally expressed as: A = Ao * e ^ (-kt) => A/Ao = e^(-kt) Half-life time => A/Ao = 1/2, and t = 4.5 min => 1/2 = e^(-k*4.5) => ln(2) = 4.5k => k = ln(2) / 4.5 ≈ 0.154. Now substituting k, Ao = 28g, and t = 7 min to determine the remaining grams of Thallium-207 gives: A = Ao e ^ (-kt) = 28 g * e ^( -0.154 * 7) = 9.5 g. Final answer is 9.5 g.
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(a) calculate the %ic of the interatomic bond for the intermetallic compound tial3. (b) on the basis of this result, what type o
Tems11 [2777]

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

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8 0
3 months ago
What features of this model will help Armando answer the question?
lions [2927]

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

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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.

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2 months ago
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