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3 years ago

When 54.50 and 50.407 are divided, how many significant figure would you have?

Chemistry

1 answer:

Anastaziya [24]3 years ago

8 0

Answer:

4 sig figs

Explanation:

<h3>Rule when dividing or multiplying:</h3>

After normally solving the expression cut round the decimal to the least accurate of the two.

54.50/50.407 = 1.08119904

Since 54.50 is the least accurate we will round to the hundredths place.

1.081 is final answer

so there are 4 sig figs.

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Nitrogen is a group 15 element. What does being In this group imply about the structure of the nitrogen atom?

sveticcg [70]

Answer:

Nitrogen has 5 valence electrons

Explanation:

nitrogen has it's attoms form triple bonds which are very hard to break

so non-reactive

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3 years ago

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Which of the following elements does not form hydride by direct heating with

ozzi

I believe the answer is A.

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2 years ago

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Why do waves move faster at higher temperatures and in a solid phase?

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<span>There is more energy and the particles are closer together.</span>

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3 years ago

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Ideal gas (n 2.388 moles) is heated at constant volume from T1 299.5 K to final temperature T2 369.5 K. Calculate the work and h

bija089 [108]

Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

Explanation :

(a) At constant volume condition the entropy change of the gas is:

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

We know that,

The relation between the $C_p\text{ and }C_v$ for an ideal gas are :

$C_p-C_v=R$

As we are given :

$C_p=28.253J/K.mole$

$28.253J/K.mole-C_v=8.314J/K.mole$

$C_v=19.939J/K.mole$

Now we have to calculate the entropy change of the gas.

$\Delta S=-n\times C_v\ln \frac{T_2}{T_1}$

$\Delta S=-2.388\times 19.939J/K.mole\ln \frac{369.5K}{299.5K}=-10J$

(b) As we know that, the work done for isochoric (constant volume) is equal to zero. $(w=-pdV)$

(C) Heat during the process will be,

$q=n\times C_v\times (T_2-T_1)=2.388mole\times 19.939J/K.mole\times (369.5-299.5)K= 3333.003J$

Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.

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3 years ago

Combining a base and an acid forms

Rainbow [258]

Answer: C. Salt and Water

Explanation:

The reaction between an acid and a base will have salt and water as its end product. This process is known " Neutralization Reaction".

HCl + NaOH -----------> NaCl + H2O

Hydrochloric Sodium Sodium Water

Acid Hydroxide (base) Chloride(Salt)

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3 years ago