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irina [24]
3 years ago
8

3. Solve the following problems. Work must be shown in order to earn credit.

Chemistry
1 answer:
madreJ [45]3 years ago
4 0
C. A container of mixed gases is 20.9 C I hope this helps
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The element gallium has two stable isotopes. About 60% of the gallium atoms found are gallium-69, while the remaining 40% are ga
Varvara68 [4.7K]

Answer:

70 amu

Explanation:

The average atomic mass of an element is calculated by first finding the products of the atomic masses of the isotopes of that element and their respective percent abundances. The second/last step is finding the sum of those products, applying significant figures rules throughout. In this case, you would do the following:

(69 x 60%) + (71 x 40%) =  70 amu (with significant figures rules applied)

Cheers

4 0
3 years ago
13. Acid-catalyzed dehydration-condensation reactions of carboxylic acids and alcohols produce chemicals called
Sliva [168]

Answer:

A condensation reaction joins two larger molecules by splitting out a smaller molecule — usually water or ammonia — between them.

In an esterification reaction, an OH  from the acid and an H from the alcohol form a molecule of water.

The larger parts join to form the ester.

Ethanol and butanoic acid react to form ethyl butanoate.

7 0
3 years ago
Need help with balance equation in chemistry
emmasim [6.3K]
The balance is A Go check other answer I posted
6 0
2 years ago
The specific gravity of iron is 7.87, and the density of water at 4.00
OlgaM077 [116]

The volume of 6.00g iron is 0.762 cm^3.

<em>Step 1</em>. Calculate the <em>density of Fe</em>

The <em>specific gravity</em> (SG) of a substance is the ratio of its density to the density of water at 4 °C.

SG = density of substance/density of water.

∴ 7.87 = density of iron/1.00 g·cm^(-3)

Density of Fe = 7.87 × 1.00 g·cm^(-3) = 7.87 g·cm^(-3)

<em>Step 2</em>. Calculate the <em>mass of Fe</em>

Volume of Fe = 6.00 g Fe × (1 cm^3 Fe/7.87 g Fe) = <em>0.762 cm^3</em>

8 0
4 years ago
A chemical compound has a molecular weight of 89.05 g/mole. 1.400 grams of this compound underwent complete combustion under con
Nataly_w [17]

Answer:

\Delta _{comb}H=-2,265\frac{kJ}{mol}

Explanation:

Hello!

In this case, for such calorimetry problem, we can notice that the combustion of the compound releases the heat which causes the increase of the temperature by 11.95 °C, it means that we can write:

Q _{comb}=-C_{calorimeter}\Delta T_{calorimeter}

In such a way, we can compute the total released heat due to the combustion considering the calorimeter specific heat and the temperature raise:

Q _{comb}=-2980\frac{J}{\°C} *11.95\°C\\\\Q _{comb}=-35,611J

Next, we compute the molar heat of combustion of the compound by dividing by the moles, considering 1.400 g were combusted:

n=1.400g*\frac{1mol}{89.05g} =0.01572mol

Thus, we obtain:

\Delta _{comb}H=\frac{Q_{comb}}{n}=\frac{-35,611J}{0.01572mol}  \\\\\Delta _{comb}H=-2,265,331\frac{J}{mol}*\frac{1kJ}{1000J}  \\\\\Delta _{comb}H=-2,265\frac{kJ}{mol}

Best regards!

7 0
3 years ago
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