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9966 [12]
3 years ago
9

Element X is found in two forms: 90.0% is an isotope that has a mass of 20.0, and 10.0% is an isotope that has a mass of 22.0. W

hat is the atomic mass of element X?
A. 20.0
B. 20.2
C. 20.8
D. 21.2
E. 21.8
Can anyone explain this one for me?
Chemistry
1 answer:
VMariaS [17]3 years ago
7 0
M_{X} = \frac{(20.0\cdot 90\%)+(22.0\cdot10\%)}{100\%} = \frac{2020}{100} = 20.2[u]

answer: B
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What is the change in enthalpy when 11. 0 g of liquid mercury is heated by 15°c?
Vlad1618 [11]

Enthalpy change is the difference between energy used and energy gained. The change in enthalpy of the liquid mercury is 0.0231 kJ.

<h3>What is the enthalpy change?</h3>

Enthalpy change is the difference between the energy used to break chemical bonds and the energy gained by the products formed in a chemical reaction.

The enthalpy change is given by,

\rm \Delta H_{rxn} = \rm q_{rxn}

and,

\rm q = mc\Delta T

Given,

Mass of the liquid mercury (m) = 11.0 gm

The specific heat of mercury (c) = 0.14 J per g per degree Celsius

Temperature change = 15 degrees Celsius

Enthalpy change is calculated as:

\begin{aligned} \rm q &= \rm mc\Delta T\\\\&= 11 \times 0.14 \times 15\\\\&= 23.1 \;\rm J\end{aligned}

Therefore, 0.0231 kJ is the change in enthalpy.

Learn more about enthalpy change here:

brainly.com/question/10932978

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6 0
2 years ago
At standard temperature and pressure. 0.500 mole of xenon gas occupies
ANEK [815]

Answer:

0.500 mole of Xe (g) occupies 11.2 L at STP.

General Formulas and Concepts:

<u>Gas Laws</u>

  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Stoichiometry</u>

  • Mole ratio
  • Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

<em>Identify.</em>

0.500 mole Xe (g)

<u>Step 2: Convert</u>

  1. [DA] Set up:                                                                                                  \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg)
  2. [DA] Evaluate:                                                                                               \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg) = 11.2 \ \text{L Xe}

Topic: AP Chemistry

Unit: Stoichiometry

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1.5?10000 10% then add it up
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lora16 [44]
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Hope this answers the question.
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