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3 years ago

True or False

Chemistry

1 answer:

serious [3.7K]3 years ago

3 0

Answer:

1. False

2. True

3. True

4. True

5. True

6. True

7. True

8. False

9. True

Explanation:

Density is a physical property since its measurement does not involve any chemical process.

Since transition elements exhibit variable oxidation states, the actual oxidation state of the transition element must be specified in the compound.

Due to the fact that neutron has no charge, it was discovered by Chadwick long after the electron and proton were discovered.

The balancing of chemical reaction equations is a demonstration that atoms are neither created no destroyed. It also shows that mass is neither created nor destroyed in chemical reactions.

When a gas is heated, it expands. Its volume and its kinetic energy increases. Since volume and pressure are inversely proportional (Boyle's law) the pressure decreases.

Enthalpy is said to be an extensive property. This implies that the magnitude of change in enthalpy is known to depend on the amount of reactants that is actually reacted.

The combined gas law is applicable to all ideal gas systems irrespective of their individual chemical formulas.

10g of CO2 contains 0.227 moles of CO2 while 10g of NO contains 0.33 moles of NO hence 10.0 g of NO will contain more molecules than 10.0g of CO2.

If a sample is not at absolute zero, the particles are known to possess kinetic energy which decreases continuously until absolute zero is attained.

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A _____ forecasts the weather using data collected from many sources.

Sav [38]

Meteorologist.............................

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3 years ago

Balance the combustion eqation; __C5H + __O2 ---> __CO2+__H2O

stira [4]

This equation C5H + O2 ---> CO2 + H2O has a mistake.

C5H is wrong. You missed the subscript of H.

I will do it for you assuming some subscript to show you the procedure, but you have to use the right equation to get the right balanced equation.

Assuming the tha combustion equation is C5H12 + O2 ---> CO2 + H2O

First you need to balance C, so you put a 5 before CO2 and get

C5H12 + O2 ---> 5CO2 + H2O

Now you count the hydrogens: 12 on the left and 2 on the right. So put a 6 before H2O and get:

C5H12 + O2 ---> 5CO2 + 6H2O

Now count the oxygens: 2 on the left and 16 on the right, so put an 8 on before O2:

=> C5H12 + 8O2 ---> 5CO2 + 6H2O.

You can verify that the equation is balanced

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3 years ago

(c) O2 gas is transferred from a 3 L vessel containing oxygen at 4 atm to an evacuated 20 L vessel at a constant temperature of

Fudgin [204]

Answer:

After the transfer the pressure inside the 20 L vessel is 0.6 atm.

Explanation:

Considering O2 as an ideal gas, it is at an initial state (1) with V1 = 3L and P1 = 4 atm. And a final state (2) with V2 = 20L. The temperature remain constant at all the process, thus here applies the Boyle-Mariotte law. This law establishes that at a constant temperature an ideal gas the relationship between pressure and volume remain constant at all time:

$P x V = k$

Therefore, for this problem the step by step explanation is:

$P_{1} xV_{1} = P_{2} xV_{2}$

Clearing P2 and replacing

$P_{2}= \frac{P_{1} xV_{1}}{V_{2} } = \frac{4atmx3L}{20L} = 0.6atm$

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3 years ago

During decomposition of water,10g of oxygen gas was formed.How much hydrogen gas was formed

statuscvo [17]

Answer:

https://www.quora.com/The-following-chemical-reaction-shows-the-decomposition-of-water-to-form-hydrogen-gas-and-oxygen-gas-2H2O-I-produces-2H2-g-O2-g-if-10-0-grams-of-water-reacted-and-you-found-1-11-grams-of-H2-formed-how-many-of-O2

Explanation: i hope ths helps you

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3 years ago

The following reaction produces ethanoic acid (CHACOOH) from methanol (CH3OH) and carbon

tigry1 [53]

Answer:

$\boxed{\text{300 g}}$

Explanation:

We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

M_r: 32 60

CH₃OH + CO ⟶ CH₃COOH

m/g: 160

(a) Moles of CH₃OH

$\text{Moles of CH$_{3}$OH} = \text{160 g CH$_{3}$OH }\times \dfrac{\text{1 mol CH$_{3}$OH }}{\text{32 g CH$_{3}$OH}}= \text{5.00 mol CH$_{3}$OH}$

(b) Moles of CH₃COOH

$\text{Moles of CH$_{3}$COOH} = \text{5.00 mol CH$_{3}$OH } \times \dfrac{\text{1 mol CH$_{3}$COOH}}{\text{1 mol CH$_{3}$OH }} = \text{5.00 mol CH$_{3}$COOH}$

(c) Mass of CH₃COOH

$\text{Mass of CH$_{3}$COOH} =\text{5.00 mol CH$_{3}$COOH} \times \dfrac{\text{60 g CH$_{3}$COOH}}{\text{1 mol CH$_{3}$COOH}} = \textbf{300 g CH$_{3}$COOH}\\\\\text{The maximum mass of ethanoic acid that can be produced is $\boxed{\textbf{300 g}}$}$

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3 years ago