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Neko [114]
1 year ago
7

6) A 0.20 ml CO2 bubble in a cake batter is at 27°C. In the oven it gets

Chemistry
1 answer:
Nata [24]1 year ago
5 0

Answer: The new volume of cake is 1.31 mL.

Explanation:

Given: V_{1} = 0.20 mL,         T_{1} = 27^{o}C

V_{2} = ?,                    T_{2} = 177^{o}C

Formula used to calculate new volume is as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}

Substitute the values into above formula as follows.

\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{0.20 mL}{27^{o}C} = \frac{V_{2}}{177^{o}C}\\V_{2} = 1.31 mL

Thus, we can conclude that the new volume of cake is 1.31 mL.

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AVprozaik [17]

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No, the puddle was formed because of the sun, because if there was snow and it rained then it would have turned slippery or icy

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Dibutyl phthalate is a plasticizer added to make brittle polymers softer and more flexible. Dibutyl phthalate can be prepared by
umka21 [38]

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Explanation:

find the solution below

4 0
2 years ago
When backpacking in the wilderness, hikers often boil water to sterilize it for drinking. Suppose that you are planning a backpa
Pavlova-9 [17]

Answer:

2.104 L fuel

Explanation:

Given that:

Volume of water = 35 L = 35 × 10³ mL

initial temperature of water = 25.0 ° C

The amount of heat needed to boil water at this temperature can be calculated by using the formula:

q_{boiling} = mc \Delta T

where

specific heat   of water c= 4.18 J/g° C

q_{boiling} = 35 \times 10^{3} \times \dfrac{1.00 \ g}{1 \ mL} \times 4.18  \ J/g^0 C \times (100 - 25)^0 C

q_{boiling} = 10.9725 \times 10^6 \ J

Also; Assume that the fuel has an average formula of C7 H16 and 15% of the heat generated from combustion goes to heat the water;

thus the heat of combustion can be determined via the expression

q_{combustion} =-  \dfrac{q_{boiling}}{0.15}

q_{combustion} =-  \dfrac{10.9725 \times 10^6 J}{0.15}

q_{combustion} = -7.315 \times 10^{7} \ J

q_{combustion} = -7.315 \times 10^{4} \ kJ

For heptane; the equation for its combustion reaction can be written as:

C_7H_{16} + 11O_{2(g)} -----> 7CO_{2(g)}+ 8H_2O_{(g)}

The standard enthalpies of the  products and the reactants are:

\Delta H _f   \ CO_{2(g)} = -393.5 kJ/mol

\Delta H _f   \ H_2O_{(g)} = -242 kJ/mol

\Delta H _f   \ C_7H_{16 }_{(g)} = -224.4 kJ/mol

\Delta H _f   \ O_{2{(g)}} = 0 kJ/mol

Therefore; the standard enthalpy for this combustion reaction is:

\Delta H ^0= \sum n_p\Delta H^0_{f(products)}- \sum n_r\Delta H^0_{f(reactants)}

\Delta H^0 =( 7  \ mol ( -393.5 \ kJ/mol)  + 8 \ mol (-242 \ kJ/mol) -1 \ mol( -224.4 \ kJ/mol) - 11  \ mol  (0 \ kJ/mol))

\Delta H^0 = (-2754.5 \ \  kJ -  1936 \ \  kJ+224.4 \  \ kJ+0 \ \  kJ)

\Delta H^0 = -4466.1 \ kJ

This simply implies that the amount of heat released from 1 mol of C7H16 = 4466.1 kJ

However the number of moles of fuel required to burn 7.315 \times 10^{4} \ kJ heat released is:

n_{fuel} = \dfrac{q}{\Delta \ H^0}

n_{fuel} = \dfrac{-7.315 \times 10^{4} \ kJ}{-4466.1  \ kJ}

n_{fuel} = 16.38  \ mol \ of \ C_7 H_{16

Since number of moles = mass/molar mass

The  mass of the fuel is:

m_{fuel } = 16.38 mol \times 100.198 \ g/mol}

m_{fuel } = 1.641 \times 10^{3} \ g

Given that the density of the fuel is = 0.78 g/mL

and we know that :

density = mass/volume

therefore making volume the subject of the formula in order to determine the volume of the fuel ; we have

volume of the fuel = mass of the fuel / density of the fuel

volume of the fuel = \dfrac{1.641 \times 10^3 \ g }{0.78  g/mL} \times \dfrac{L}{10^3 \ mL}

volume of the fuel  = 2.104 L fuel

3 0
2 years ago
A sample of copper with a mass of 63.5g contains 6.02 x10^23 atoms calculate the mass of an average copper atom
m_a_m_a [10]

Answer:

The mass of an average copper atom is 1.0548\times 10^{-22}\ g

Explanation:

Given:

The total mass of copper atoms, m = 63.5\ g

Number of atoms, N=6.02\times 10^{23}

Now, we are asked to find the mass of 1 copper atom.

We use unitary method to find the mass of 1 copper atom.

Mass of N atoms = m

∴ Mass of 1 atom = \frac{m}{N}

Plug in 63.5 for 'm', 6.02\times 10^{23} for 'N' and simply.

Mass of 1 atom = \dfrac{63.5}{6.02\times 10^{23}}=1.0548\times 10^{-22}\ g

Therefore, the mass of an average copper atom is 1.0548\times 10^{-22}\ g

5 0
2 years ago
Rank the boiling points of the following molecules from highest to lowest. butanone diethyl ether butane and butanol.
Gekata [30.6K]

Answer:

From highest to lowest:

butanol: 117.7 degree Celsius

butanone: 79.64 degree Celsius

diethyl ether: 34.6 degree Celsius

n-butane: -0.4 degree Celsius

7 0
1 year ago
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