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dolphi86 [110]
4 years ago
7

Compute 4.659×104−2.14×104. Round the answer appropriately.

Chemistry
1 answer:
antiseptic1488 [7]4 years ago
7 0
<span>The answer is 2.519 × 10^4. Use the distributive property: a × x + b × x = (a + b) × x. If a = 4.659, b = 2.14, and x = 10^4, then 4.659 × 10^4 − 2.14 × 10^4 = (4.659 - 2.14) × 10^4. Now, subtract numbers in parenthesis: 4.659 × 10^4 − 2.14 × 10^4 = (4.659 - 2.14) × 10^4 = 2.519 × 10^4.Hope this helps. Let me know if you need additional help!</span>
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Answer:

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Answer: The empirical formula is C_3H_3O.

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We are given:

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For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 12.24 g of carbon dioxide, =\frac{12}{44}\times 12.24=3.338g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 2.505 g of water, =\frac{2}{18}\times 2.505=0.278g of hydrogen will be contained.

Mass of oxygen in the compound = (5.287) - (3.338+0.278) = 1.671  g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{3.338g}{12g/mole}=0.278moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.278g}{1g/mole}=0.278moles

Moles of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{1.671g}{16g/mole}=0.104moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.278}{0.104}=3

For H =\frac{0.278}{0.104}=3

For O =\frac{0.104}{0.104}=1

The ratio of C : H : O = 3: 3: 1

Hence the empirical formula is C_3H_3O.

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