How many atoms are in 2.00 moles of Cu?

If the same amount of heat is added to 25.0 g of each of the metals, which are all at the same initial temperature, which metal

AVprozaik [17]

Answer:

The bismuth sample.

Explanation:

The specific heat $c$ of a substance (might not be a metal) is the amount of heat required for heating a unit mass of this substance by unit temperature (e.g., $\rm 1\; ^{\circ}C$ .) The formula for specific heat is:

$\displaystyle c = \frac{Q}{m \cdot \Delta T}$ ,

where

$Q$ is the amount of heat supplied.
$m$ is the mass of the sample.
$\Delta T$ is the increase in temperature.

In this question, the value of $Q$ (amount of heat supplied to the metal) and $m$ (mass of the metal sample) are the same for all four metals. To find $\Delta T$ (change in temperature,) rearrange the equation:

$\displaystyle c \cdot \Delta T = \frac{Q}{m}$ ,

$\displaystyle \Delta T = \frac{Q}{c \cdot m}$ .

In other words, the change in temperature of the sample, $\Delta T$ can be expressed as a fraction. Additionally, the specific heat of sample, $c$ , is in the denominator of that fraction. Hence, the value of the fraction would be the largest for sample with the smallest specific heat.

Make sure that all the specific heat values are in the same unit. Find the one with the smallest specific heat: bismuth ( $\rm 0.123 \; J \cdot g\cdot \,^{\circ}C^{-1}$ .) That sample would have the greatest increase in temperature. Since all six samples started at the same temperature, the bismuth sample would also have the highest final temperature.

3 0

2 years ago

If I had 3.50 x 10 24molecules of Cl2 gas, how many grams is this?

Zinaida [17]

Answer:

412 g Cl₂

General Formulas and Concepts:

Atomic Structure

Reading a Periodic Table
Moles
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 3.50 × 10²⁴ molecules Cl₂

[Solve] grams Cl₂

Step 2: Identify Conversions

Avogadro's Number

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of Cl₂ - 2(35.45) = 70.9 g/mol

Step 3: Convert

[DA] Set up: $\displaystyle 3.50 \cdot 10^{24} \ molecules \ Cl_2(\frac{1 \ mol \ Cl_2}{6.022 \cdot 10^{23} \ molecules \ Cl_2})(\frac{70.9 \ g \ Cl_2}{1 \ mol \ Cl_2})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 412.072 \ g \ Cl_2$