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3 years ago

Is cold water more dense

1 answer:

5 0

Cold water is more dense and will sink in room-temperature water. hope it helps :)

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What makes atoms different than ions?

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Answer:the term 'atom' in chemistry represents the basic unit of matter that normally exists in a free state and contains all chemical properties. It is an infinite particle that clearly identifies a chemical element. It consists of a positively charged nucleus and is surrounded by negatively charged electrons.

An ion or charge is an atom or group of atoms that has lost or gained one or more electrons. An ion that has a negative charge, when it gains one or more electrons, is called an anion or negative charge, and an ion that has a positive charge when it loses one or more electrons, is called a cation or positive charge.

Explanation:

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a lab is performed to do the study of pH and heat on the function of the catalase enzyme found in carrot plants. what would be c

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It would be the ph and heat

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What quantity measures the amount of space an object occupies?

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The correct one would be volume

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In a healthy pond, the temperature is 16°C (61°F). What is the most likely pH of this pond?

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A . 6.3 In a healthy pond, the temperature is 16°C (61°F). What is the most likely pH of this pond

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3 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