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Chemistry

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crimeas [40]2 years ago

8 0

Answer:

Answer is A Peasants were valued members of society in Japan but not in Europe.

Explanation:

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The rock outcrop, shown above demonstrates the law of included fragments. Identify the correct definition for the law of include

Radda [10]

Answer:

Explanation:

The law of included fragments states: Fragments found in a rock must be older than the rock itself.

7 0

2 years ago

A scientist fills a large, tightly sealed balloon with 75,000 mL of helium at STP.

pav-90 [236]

Considering the definition of STP conditions, 3.35 moles of helium are contained within the balloon.

<h3>Definition of STP condition</h3>

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

<h3>Amount of moles of helium within the balloon</h3>

In this case, you know that scientist fills a large, tightly sealed balloon with 75,000 mL of helium at STP.

So, you can apply the following rule of three: if by definition of STP conditions 22.4 liters are occupied by 1 mole of helium, 75 L (75 L= 75000 mL, being 1 L= 1000 L) are occupied by how many moles of helium?

$amount of moles of helium=\frac{75 Lx1 mole}{22.4 L}$

<u><em>amount of moles of helium= 3.35 moles</em></u>

Finally, 3.35 moles of helium are contained within the balloon.

Learn more about STP conditions:

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3 0

2 years ago

The standard length of a certain golf club is 116.84 cm. You measure the club to be 116.85 cm long. You measure it again and fin

IgorC [24]

Answer:

The first measurement is 116.84 cm, and the second measurement is 116.85 cm.

Explanation:

The problem provides both the first and second measurements so all you have to do is fill in the blank with them.

8 0

3 years ago

Consider the hypothetical reaction 4A + 3B → C + 2D Over an interval of 3.00 s the average rate of change of the concentration o

pogonyaev

Answer:

Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M

Explanation:

4A + 3B ------> C + 2D

In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s

The amount of A that has reacted at the end of 3 seconds will be

0.08 × 3 = 0.24 M

Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.

From the chemical reaction,

4 moles of A gives 1 mole of C

0.24 M of reacted A will form (0.24 × 1)/4 M of C

Amount of C formed at the end of the 3s interval = 0.06 M

If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.

If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M

3 0

3 years ago

The map below shows the range of three types of grasslands in the United

Sav [38]

I’m so sorry about the other guy! The answer is A. Infertile soil honey

God Bless!! <3

6 0

2 years ago

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