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3 years ago

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How many significant figures are in the measurement 603.040 g

1 answer:

Vlad [161]3 years ago

3 0

603.040 has 5 significant figures in it I'm pretty sure

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For a reaction in a galvanic cell both ΔH° and ΔS° are positive. Which of the following statements is true?

irina [24]

Answer:

Option C is the correct statement.

ℰ°cell will increase with an increase in temperature

Explanation:

To understand this problem it is neccessarry to state the relatioonships between the various paramenters mentioned in the question. these parameters include ΔH°, ΔS°, ΔG° and ℰ°Cell.

ΔH°, ΔS°, ΔG° are related with the following equation;

ΔG° = ΔH° - TΔS°

From the above equation we see that ΔG° cannot be greater than 0 for all temperatures. At higher temperatures, ΔG° becomes a negative value.

A spontaneous redox reaction is characterized by a negative value of ΔG and a positive value of ℰ°Cell, consistent with our earlier discussions. When both reactants and products are in their standard states, the relationship between ΔG° and ℰ°Cell is as follows:

ΔG∘=−nFℰ°Cell where;

F = Faraday and n = number of moles

Since ΔG° becomes negative at higher temperatures, it means ℰ°Cell would increase as temperature increases.

The relationship between G, Ecell and temperature can be given as;

Increase in Temperature = Negative value of G = Increase in E

Due to the fact established above; options B, D and E are false. This means the correct option is option C - ℰ°cell will increase with an increase in temperature

3 0

3 years ago

What is the boiling point of a solution that contains 3 moles of KBr in 2000 g of water?

bija089 [108]

Answer:

C. 101.5°C

Explanation:

The computation of the boiling point of a solution is shown below:

As we know that

$\Delta T = i K_bm$

where,

i = Factor of Van’t Hoff i.e 2KBr

K_b = Boiling point i.e 0.512° C

m = molality $= \frac{Moles\ of\ solute}{Solvent\ mass}$

Now place the values to the above formula

So, the boiling point is

$(T - 100^\circ C) = 2\times 0.512^\circ \times \frac{3.00\ ml}{2.00\ kg}$

After solving this, the t is

= 100° C + 1.536° C

= 101.536° C

Hence, the correct option is C.

5 0

3 years ago

with strong heating calcium carbonate undergoes thermal decomposition how many mole of CaCO3 are there in 50g of calcium carbona

aev [14]

Ca=40
C=12
O=16
1 mole of CaCO3 has 100 grams
So 50 grams is 0.5 mole

6 0

2 years ago

Read 2 more answers

A neutron collides with a nitrogen atom, resulting in a transmutation. Balance the equation. Superscript 1 Subscript 0 Baseline

o-na [289]

Answer:

A is 14

Z is 6

X is C

Explanation:

Have a great day!

7 0

3 years ago

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

Read 2 more answers