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jasenka [17]
3 years ago
5

Ethan is observing chemical and physical properties of a substance. He heats a substance and observes that the substance turns f

rom a brown solid to a black powder. He refers to several chemistry journals that claim this represents a chemical reaction. From his observation and research, he concludes that the substance goes through a chemical change when heated. How can Ethan best defend his conclusion?
Chemistry
1 answer:
densk [106]3 years ago
3 0

Explanation:

A chemical reaction is defined as the reaction which leads to change in chemical composition of the reactants due to which there occurs formation of a new compound.

For example, when Ethan heats a substance and observes that the substance turns from a brown solid to a black powder then it means heating has lead to a chemical change in the brown solid because of which it has turned into a black powder.

Whereas if there occurs no change in chemical composition of the reactants then we can say that a physical change has occurred.

For example, change in shape, size, mass etc of a substance are all physical changes.

Hence, we can conclude that Ethan best defend his conclusion as a chemical change.

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What type of climate conditions are needed to form coal deposits.
guajiro [1.7K]

Answer:

Subbituminous coal can form at temperatures as low as 35 to 80 °C (95 to 176 °F) while anthracite requires a temperature of at least 180 to 245 °C (356 to 473 °F).

Sub-types: Cannel coal

Child material class: Lignite

Explanation:

6 0
2 years ago
How much heat is lost when 0.440 mol of steam condenses at 100 °C?
dusya [7]

Answer:

17,890 J

Explanation:

The amount of heat released by a gaseous substance when it condenses is given by the formula

Q=n\lambda_v

where

n is the number of moles of the substance

\lambda_v is the latent heat of vaporization

The formula can be applied if the substance is at its vaporization temperature.

In this problem, we have:

n = 0.440 mol is the number of moles of steam

\lambda_v=40,660 J/mol is the latent heat of vaporization of water

And the steam is already at 100C, so we can apply the formula:

Q=(0.440)(40660)=17,890 J

8 0
3 years ago
Burning 12.00 g of an oxoacid produces 17.95 g of carbon dioxide and 4.87 g of water. Consider that 0.25
Veseljchak [2.6K]

Answer: The molecular formula will be C_6H_6O_6

Explanation:

Mass of CO_2 = 17.95 g

Mass of H_2O= 4.87 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 17.95 g of carbon dioxide, =\frac{12}{44}\times 17.95=4.89g of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.87 g of water, =\frac{2}{18}\times 4.87=0.541g of hydrogen will be contained.

Mass of oxygen in the compound = (12.00) - (4.89+0.541) = 6.57 g

Mass of C = 4.89 g

Mass of H =  0.541 g

Mass of O = 6.57 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{4.89g}{12g/mole}=0.407moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.541g}{1g/mole}=0.541moles

Moles of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{6.57g}{16g/mole}=0.410moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.407}{0.407}=1

For H =\frac{0.541}{0.407}=1

For O = \frac{0.410}{0.407}=1

The ratio of C : H : O = 1: 1  : 1

Hence the empirical formula is CHO.

Hence the empirical formula is CHO

The empirical weight of CHO = 1(12)+1(1)+1(16)= 29 g.

If 0.25 moles has mass of 44.0 g

Thus 1 mole has mass of = \frac{44.0}{0.25}\times 1=176g

Thus molecular mass is 176 g

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{176g}{29g}=6

The molecular formula will be=6\times CHO=C_6H_6O_6

5 0
3 years ago
Defination of hydrocarbons
sergey [27]
Compound made up of hydrogen and carbon
7 0
2 years ago
"46.7 g of water at 80.6 oC is added to a calorimeter that contains 45.33 g of water at 40.6 oC. If the final temperature of the
soldier1979 [14.2K]

<u>Answer:</u> The specific heat of calorimeter is 30.68 J/g°C

<u>Explanation:</u>

When hot water is added to the calorimeter, the amount of heat released by the hot water will be equal to the amount of heat absorbed by cold water and calorimeter.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[(m_2\times c_2)+c_3](T_{final}-T_2)       ......(1)

where,

q = heat absorbed or released

m_1 = mass of hot water = 46.7 g

m_2 = mass of cold water = 45.33 g

T_{final} = final temperature = 59.4°C

T_1 = initial temperature of hot water = 80.6°C

T_2 = initial temperature of cold water = 40.6°C

c_1 = specific heat of hot water = 4.184 J/g°C

c_2 = specific heat of cold water = 4.184 J/g°C

c_3 = specific heat of calorimeter = ? J/g°C

Putting values in equation 1, we get:

46.7\times 4.184\times (59.4-80.6)=-[(45.33\times 4.184)+c_3](59.4-40.6)

c_3=30.68J/g^oC

Hence, the specific heat of calorimeter is 30.68 J/g°C

6 0
3 years ago
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