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2 years ago

Which metal will gain heat faster. Brass or Gold?

2 answers:

6 0

Gold will gain heat faster because it's denser than Brass or some other metals. When an object has a high density it will actually gain heat QUICKER than low density objects.

FinnZ [79.3K]2 years ago

5 0

Hey!!

here is your answer >>>

In order to find out which metal gains heat faster, we will have to find out their specific heat!. Specific heat is the heat required to raise the temperature of the unit substance, The specific heat of gold and brass are 0.126 for gold and 0.380 for brass!. The specific heat of gold is less than brass!. So, gold will gain heat faster!.

Hope my answer helps!

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Please help me with this question. What are two elements that will react similarly to beryllium ?

LUCKY_DIMON [66]

Answer:

magnesium and aluminium

Explanation:

google

7 0

2 years ago

Read 2 more answers

How many molecules OF2 would have a mass of .132 g

BlackZzzverrR [31]

<h3>Answer:</h3>

1.47 × 10²¹ molecules OF₂

<h3>General Formulas and Concepts: </h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

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

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation: </h3>

<u>Step 1: Define</u>

0.132 g OF₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of O - 16.00 g/mol

Molar Mass of F - 19.00 g/mol

Molar Mass of OF₂ - 16.00 + 2(19.00) = 54.00 g/mol

<u>Step 3: Convert</u>

Set up: $\displaystyle 0.132 \ g \ OF_2(\frac{1 \ mol \ OF_2}{54.00 \ g \ OF_2})(\frac{6.022 \cdot 10^{23} \ molecules \ OF_2}{1 \ mol \ OF_2})$
Divide/Multiply: $\displaystyle 1.47204 \cdot 10^{21} \ molecules \ OF_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.47204 × 10²¹ molecules OF₂ ≈ 1.47 × 10²¹ molecules OF₂

3 0

2 years ago

A sealed container can hold

svp [43]

0.216 moles of gas can the container hold if a sealed container can hold 0.325 L of gas at 1.00 atm and 293 K.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

R = gas constant = 0.08206 L.atm / mol K

T = temperature, Kelvin

V=5 L

P = 1.05 atm

T = 296 K

Putting value in the given equation:

$\frac{PV}{RT}=n$