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3 years ago

The substances that are chemically bound together are

Chemistry

1 answer:

pshichka [43]3 years ago

4 0

Answer:

Select one of the readings in this unit and in at least one hundred words, discuss how immediacy, foreshadowing, and commentary work to show author's purpose, meaning, or tone and style.

Explanation:

Select one of the readings in this unit and in at least one hundred words, discuss how immediacy, foreshadowing, and commentary work to show author's purpose, meaning, or tone and style.

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Two moles of magnesium (Mg) and five moles of oxygen (O2) are placed in a reaction vessel. When magnesium is ignited, it reacts

Y_Kistochka [10]

Answer:

Explanation:

Two moles of magnesium (Mg) and five moles of oxygen (O2) are placed in a reaction vessel. When magnesium is ignited, it reacts with oxygen. What is the limiting reactant in this experiment?

Mg + O2 → MgO (unbalanced)

first, balance the equation

2Mg +O2-------> 2MgO

two magnesium atoms react with one diatomic oxygen molecule

there is a 1:1 ratio of magnesium to oxygen atoms

but we have 2 moles of magnesium atoms and 2X5 = 10 moles of oxygen atoms

the lesser magnesium LIMITS the amount of product we can make, so it is the LIMITING REAGENT.

6 0

2 years ago

What is the percent by mass of 15g of sugar of 70g of water?

Mama L [17]

Answer:

lol u have mrs carnes

Explanation:

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3 years ago

How many covalent bonds are predicted for each atom: (a) F; (b) Si; (c) Br; (d) O; (e) P; (f) S?

evablogger [386]

The <span>covalent bonds are predicted for each atom are :

</span>(a)F = 1
(b) Si = 4
(c) Br = 1
(d) O = 2
(e) P = 3
(f) S = 2

4 0

3 years ago

Read 2 more answers

Describe Write your own caption for this photo.

koban [17]

Answer:

Sit by the fire to warm up

Explanation:

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2 years ago

A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio

Pavlova-9 [17]

Answer:

$C_7H_6O_2$

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

$n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC$

And the moles of hydrogen due to the produced 1.50 grams of water:

$n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH$

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

$m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO$

Thus, the subscripts in the empirical formula are:

$C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2$

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

$n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol$

Which are equal to the moles of the acid:

$n_{acid}=0.0279mol$

And the molar mass:

$MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol$

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

$C_7H_6O_2$

Best regards!

8 0

3 years ago