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boyakko [2]
2 years ago
9

How many atoms are there in 5.20 mil of hafnium. Please help due in 1hr

Chemistry
1 answer:
Levart [38]2 years ago
8 0
In 5.70 mol of Hafnium there are 34,326
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A sample of hydrogen gas is found to be 8.50 L at a temperature of 38.0 C and a pressure of 725mm Hg. Find the volume at STP.
Vsevolod [243]
Temp must be Kelvin
38 C = <span> <span> <span> 311.15 </span> </span> </span> K
Volume at STP = 8.50 liters * (273.15 / 311.15) * (725 / 760) =
<span> <span> <span> 7.1182746306 </span> </span> </span> Liters

The formula to use is:

Volume at STP = Present Volume * (273.15 / Present Temp °K) * (Present Pressure (Torr) / 760)

6 0
3 years ago
Help me with this this is something i dont know and its not on here please help meeeeeee
denis-greek [22]

Answer:

it would be option A

Explanation:

This is becuase if you look at the chart you can see tyhat the group of rats that got feed to vitamans did gain more wati then the ones on the normal diet.

8 0
2 years ago
When magnesium combines with oxygen, the reaction involves
nikdorinn [45]

Answer:

The correct answer is option c. transfer of electrons from Mg to O.

Explanation:

Hello!

Let's solve this!

When Magnesium (Mg) reacts with oxygen (O), magnesium oxide is formed.

This reaction is spontaneous and occurs with oxidation number +2 of magnesium and oxidation number -2 of oxygen. It is an ionic union, so magnesium transfers its electrons to oxygen.

We conclude that the correct answer is option c. transfer of electrons from Mg to O.

7 0
3 years ago
One question please help!
Agata [3.3K]
<span>1 trial : you have nothing to compare the result with - you don't know if it's a mistake.
2 trials : you can compare results - if very different, one may have gone wrong, but which one?
3 trials : if 2 results are close and 3rd far away, 3rd probably unreliable and can be rejected.

******************************

First calculate the enthalpy of fusion. M, C and m,c = mass and specific heat of calorimeter and water; n, L = mass and heat of fusion of ice; T = temperature fall.

L = (mc+MC)T/n.

c=4.18 J/gK. I assume calorimeter was copper, so C=0.385 J/gK.

1. M = 409g, m = 45g. T = 22c, n = 14g
L = (45*4.18+409*0.385)*22/14 = 543.0 J/g.

2. M = 409g, m = 49g, T = 20c, n = 13g
L = (49*4.18+409*0.385)*20/13 = 557.4 J/g.

3. M = 409g, m = 54g, T = 20c, n = 14g
L = (54*4.18+409*0.385)*20/14 = 547.4 J/g.

(i) Estimate error in L from spread of 3 results.
Average L = 549.3 J/g.
average of squared differences (variance) = (6.236^2+8.095^2+1.859^2)/3 = 35.96
standard deviation = 5.9964
standard error = SD/(N-1) = 5.9964/2 = 3 J/g approx.

% error = 3/547 x 100% = 0.5%.

(ii) Estimate error in L from accuracy of measurements:
error in masses = +/-0.5g
error in T = +/-0.5c

For Trial 3
M = 409g, error = 0.5g
m = 463-409, error = sqrt(0.5^2+0.5^2) = 0.5*sqrt(2)
n =(516-463)-(448-409)=14, error = 0.5*sqrt(4) = 1.0g
K = (mc+MC)=383, error = sqrt[2*(0.5*4.18)^2+(0.5*0.385)^2] = 2.962

L = K*T/n
% errors are
K: 3/383 x 100% = 0.77
T: 0.5/20 x 100% = 2.5
n: 1.0/14 x 100% = 7.14

% errors in K and T are << error in n, so we can ignore them.
% error in L = same as in n = 7% x 547.4 = 40 (always round final error to 1 sig fig).

*************************************

The result is (i) L= 549 +/- 3 J/g or (ii) L = 550 +/- 40 J/g.
Both are very far above accepted figure of 334 J/g, so there is at least one systematic error in the experiment or the calculations.
eg calorimeter may not be copper, so C is not 0.385 J/gK. (If it was polystyrene, which absorbs/ transmits little heat, the effective value of C would be very low, reducing L.)
Using +/- 40 is probably best (more cautious).
However, the spread in the actual results is much smaller; try to explain this discrepancy - eg
* measurements were "fiddled" to get better results; other Trials were made but only best 3 were chosen.
* measurements were more accurate than I assumed (eg masses to nearest 0.1g but rounded to 1g when written down).

Other sources of error:
L=(mc+MC)T/n is too high, so n (ice melted) may be too small, or T (temp fall) too high - why?
* it is suspicious that all final temperatures were 0c - was this actually measured or just guessed? a higher final temp would reduce L.
* we have assumed initial and final temperature of ice was 0c, it may actually have been colder, so less ice would melt - this could explain small values of n
* some water might have been left in container when unmelted ice was weighed (eg clinging to ice) - again this could explain small n;
* poor insulation - heat gained from surroundings, melting more ice, increasing n - but this would reduce measured L below 334 J/g not increase it.
* calorimeter still cold from last trial when next one started, not given time to reach same temperature as water - this would reduce n.
Hope This Helps :)
</span>
3 0
3 years ago
The atomic mass of gold is 197 g/mole, and the density of gold is 19.3 g/cm3. using this information along with the measured val
olganol [36]
The speed of sound is given by the formula
Speed of sound = square root (Young's Modulus/density)
Young's modulus for gold = 78 GPa which is 7.8 * 10^10 Pa
Speed of sound = (7.8 * 10^10 / 19.3 ) = sqrt (4.04 * 10^9 )
Speed of sound = 6.36 * 10^4 m/s <<<=== answer.
6 0
3 years ago
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