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4 years ago

How does the amount of matter change when water changes state?

Chemistry

1 answer:

Serjik [45]4 years ago

5 0

Answer:

The amount of matter will be the same - matter cannot be destroyed or created.

Explanation:

The state of matter could change, however, e.g. solid to liquid.

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If all of the energy from burning 281.0 g of propane (ΔHcomb,C3H8 = –2220 kJ/mol) is used to heat water, how many liters of wate

lapo4ka [179]

This problem is providing us with the mass of propane, its enthalpy of combustion, and the initial and final temperature of water that can be heated from the burning of this fuel. At the end, the result turns out to be 42.27 L.

<h3>Combustion:</h3>

In chemistry, combustion reactions are based on the burning of fuels by using oxygen and producing both carbon dioxide and water. For propane, we will have:

$C_3H_8+5O_2\rightarrow 3CO_2+4H_2O$

Hence, we can calculate the heat released from this reaction by using the mass, which has to be converted to moles, and the given enthalpy of combustion:

$Q=281.0g*\frac{1mol}{44.09g}*-2220\frac{kJ}{mol}*\frac{1000J}{1kJ}\\ \\ Q=-1.415x10^7 J$

<h3>Calorimetry:</h3>

In chemistry, we can analyze the mass-specific heat-temperature-heat relationship via the most general heat equation:

$Q=mC\Delta T$

Thus, since Q was obtained from the previous problem, but the sign change because the released heat is now absorbed by the water, one can calculate the mass of water that rises from 20.0°C to 100.0°C with this heat:

$m=\frac{Q}{C\Delta T} =\frac{1.415x10^7J}{4.184\frac{J}{g\°C}(100.0\°C-20.0\°C)}\\ \\m=4.227x10^4g$

Finally, we convert it to liters as required:

$V=4.227 x10^4g*\frac{1mL}{1.00g}*\frac{1L}{1000mL} \\\\V=42.27L$

Learn more about calorimetry: brainly.com/question/1407669

4 0

2 years ago

Consider the following reaction, equilibrium concentrations, and equilibrium constant at

REY [17]

Answer:

Equilibrium concentration of $H_{2}O$ is 12.5 M

Explanation:

Given reaction: $C_{2}H_{4}+H_{2}O\rightleftharpoons C_{2}H_{5}OH$

Here, $K_{c}=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}][H_{2}O]}$

where $K_{c}$ represents equilibrium constant in terms of concentration and species inside third bracket represent equilibrium concentrations

Here, $[C_{2}H_{4}]=0.015M$ , $[C_{2}H_{5}OH]=1.69M$ and $K_{c}=9.0$

So, $[H_{2}O]=\frac{[C_{2}H_{5}OH]}{[C_{2}H_{4}]\times K_{c}}=\frac{1.69}{0.015\times 9.0}=12.5M$

Hence equilibrium concentration of $H_{2}O$ is 12.5 M

5 0

3 years ago

Balance the following equation by inspection: <br> Cu + HNO3 --> Cu(NO3)2 + H2O +NO

pashok25 [27]

Answer:

f is this

Explanation:

7 0

4 years ago

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Which of the following is NOT correct about elements?

Lynna [10]

Answer: B. Elements are represented by chemical formulas.

Elements are pure substances, which means that they cannot be broken down into simpler substances. The element is the most basic substance that exists, breaking it down further means breaking it down into protons, neutrons, and electrons, which is no longer a substance.

Elements have chemical properties that allow them to form different types of bonds with other elements.

However, elements *alone* are not represented by their chemical formulas. Only chemical bonds or ions are represented by a chemical formula.

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4 years ago

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What is the OH- in a solution with a pOH of 5.17

harina [27]

Answer:

1 oxygen 1 hydrogen hydrogen peroxide I think

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3 years ago