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Vinil7 [7]
3 years ago
8

A piece of iron metal is heated to 155 degrees C and placed into a calorimeter that contains 50.0 mL of water at 18.7 degrees C.

The temperature of the water rises to 26.4 degrees C. How much heat was released by the iron?
A-1610 J
B-5520 J
C-385 J
D-2250 J
Chemistry
2 answers:
zysi [14]3 years ago
3 0
Jsjsjsdjdjkdskkeekdks prbably awnser a
Idk3 years ago
0 0

It’s A i just don’t know how. But it’s fs A.

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A synthesis reaction takes place when carbon monoxide (CO) and hydrogen gas (H2) react to form methanol (CH3OH). How many grams
jek_recluse [69]

The mass of methanol produced is 8.0 g.

We have the masses of two reactants, so this is a <em>limiting reactant</em> problem.

We know that we will need a <em>balanced equation</em> with masses, moles, and molar masses of the compounds involved.

<em>Step 1</em>. <em>Gather all the informatio</em>n in one place with molar masses above the formulas and everything else below them.

MM: ___28.01  2.016 ___32.04

_______CO + 2H_2 → CH_3OH

Mass/g: 7.0 __2.5

<em>Step 2</em>. Calculate the <em>moles of each reactant</em>

Moles of CO = 7.0 g CO × (1 mol CO/28.01g CO) = 0.250 mol CO

Moles of H_2 =2.5 g H_2 × (1 mol H_2/2.016 g H_2) = 1.24 mol H_2

<em>Step 3. </em>Identify the<em> limiting reactan</em>t

Calculate the <em>moles of CH_3OH</em> we can obtain from each reactant.

<em>From CO</em>: Moles of CH_3OH = 0.250 mol CO  × (1 mol CH_3OH /1 mol CO)

= 0.250 mol CH_3OH

<em>From H_2</em>: Moles of CH_3OH = 1.24 mol H_2 × (1 mol CH_3OH /2 mol H_2)

= 0.620 mol CH_3OH

<em>CO is the limiting reactant</em> because it gives the smaller amount of CH_3OH.

<em>Step 4</em>. Calculate the <em>mass of CH_3OH</em>

Mass of CH_3OH = 0.250 mol CH_3OH × (32.04 g CH_3OH /1 mol CH_3OH) = 8.0 g CH_3OH

3 0
3 years ago
Where would sound travel the slowest
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Wood

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It is because sound is a longitudinal wave

6 0
3 years ago
The CRC Handbook, a large reference book of chemical and physical data, lists two isotopes of indium (). The atomic mass of 4.28
ivanzaharov [21]

<u>Answer:</u> The mass of second isotope of indium is 114.904 amu

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

  .....(1)

Let the mass of isotope 2 of indium be 'x'

  • <u>For isotope 1:</u>

Mass of isotope 1 = 112.904 amu

Percentage abundance of isotope 1 = 4.28 %

Fractional abundance of isotope 1 = 0.0428

  • <u>For isotope 2:</u>

Mass of isotope 2 = x amu

Percentage abundance of isotope 2 = [100 - 4.28] = 95.72 %

Fractional abundance of isotope 2 = 0.9572

Average atomic mass of indium = 114.818 amu

Putting values in equation 1, we get:

114.818=[(112.904\times 0.0428)+(x\times 0.9572)]\\\\x=114.904amu

Hence, the mass of second isotope of indium is 114.904 amu

4 0
3 years ago
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