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djyliett [7]
3 years ago
10

Consider the following equivalent expressions: 38÷1312 and 38?ab What are the values of a and b? Enter the value of a followed b

y the value of b, separated by a comma.

Chemistry
1 answer:
Anna007 [38]3 years ago
3 0

Answer:

a = 12

b = 13

Explanation:

Consider the following equivalent expressions: 38÷1312 and 38?ab What are the values of a and b? Enter the value of a followed by the value of b, separated by a comma.

The solution is attached.

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What is the smallest unit of an element that retains the properties of that element?
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The smallest unit which maintains an element's properties is an atom.

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Phthalates used as plasticizers in rubber and plastic products are believed to act as hormone mimics in humans. The value of ΔHc
jeka94

Answer:

T_f = 25.05°C

Explanation:

Given:

the value of ΔHcomb (heat of combustion) for dimethylphthalate (C10H10O4) is = 4685 kJ/mol.

mass = 0.905g of dimethylphthalate

molar mass = 194.18g dimethylphthalate

number of moles of dimethylphthalate = ???

T_i = 21.5°C

C_{calorimeter} = 6.15 kJ/°C

T_f = ???

since we have our molar mass and mass of dimethylphthalate ;we can determine the number of moles as;

0.905g of dimethylphthalate ×  \frac{1 mole (dimethylphthalate)}{194.184g(dimethylphthalate)}

number of moles of dimethylphthalate = 0.000466 moles

Heat released = moles of dimethylphthalate × heat of combustion

=  0.000466 moles × 4685 kJ

= 21.84 kJ

∴ Heat absorbed by the calorimeter =  C_{calorimeter} (T_f-T_i} )

21.84 kJ =6.15 kJ/°C * (T_f-21,5^0C)

21.84 KJ = (6.15 kJ/^0C * T_f) - (6.15 kJ/^0C*21.5^0C)

21.84 KJ = (6.15 kJ/^0C * T_f) - 132.225 kJ

21.84 KJ + 132.225 kJ = (6.15 kJ/^0C * T_f)

154.065 kJ = (6.15 kJ/^0C * T_f)

T_f = \frac{154.065kJ}{6.15kJ/^0C}

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3 years ago
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What is the mass of a sample of metal that is heated from 58.8°C to 88.9°C with a
Vadim26 [7]

Answer:

\boxed {\boxed {\sf 333 \ grams}}

Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

Q= mc \Delta T

The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

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The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.

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Now we know three variables:

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Substitute these values into the formula.

4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)

Multiply on the right side of the equation. The units of degrees Celsius cancel.

4500.0 \ J = m (13.52694 J/g)

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}

The units of Joules cancel.

\frac {4500.0 \ J }{13.52694 J/g}= m

332.6694729 \ g =m

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

333 \ g \approx m

The mass of the sample of metal is approximately <u>333 grams.</u>

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