Answer:
It depends on the number of significant figures you are changing to
Explanation:
6.02×1023=6158.46
1 sig fig = 6000
2 sig fig = 6200
3 sig fig = 6160
4 sig fig = 6158
5 sig fig = 6158.5
6 sig fig = 6158.46
When solving significant figures you have to consider the number after each number like in the case of changing to two sig fig the number following one is five and when the number is up to or greater than five you add a value of one to the number before it. But in a case where the number is less than five you just leave it like that like in the case of changing to one sig fig
<u>Answer:</u> The amount of heat released when 0.211 moles of
reacts is 554.8 kJ
<u>Explanation:</u>
The chemical equation for the reaction of
with oxygen gas follows:

The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(5\times \Delta H_f_{(B_2O_3(s))})+(9\times \Delta H_f_{(H_2O(l))})]-[(2\times \Delta H_f_{(B_5H_9(l))})+(12\times \Delta H_f_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%285%5Ctimes%20%5CDelta%20H_f_%7B%28B_2O_3%28s%29%29%7D%29%2B%289%5Ctimes%20%5CDelta%20H_f_%7B%28H_2O%28l%29%29%7D%29%5D-%5B%282%5Ctimes%20%5CDelta%20H_f_%7B%28B_5H_9%28l%29%29%7D%29%2B%2812%5Ctimes%20%5CDelta%20H_f_%7B%28O_2%28g%29%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta H_{rxn}=[(2\times (-1272))+(9\times (-285.4))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H_{rxn}=-5259kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%282%5Ctimes%20%28-1272%29%29%2B%289%5Ctimes%20%28-285.4%29%29%5D-%5B%282%5Ctimes%20%2873.2%29%29%2B%2812%5Ctimes%20%280%29%29%5D%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D-5259kJ)
To calculate the amount of heat released for the given amount of
, we use unitary method, we get:
When 2 moles of
reacts, the amount of heat released is 5259 kJ
So, when 0.211 moles of
will react, the amount of heat released will be = 
Hence, the amount of heat released when 0.211 moles of
reacts is 554.8 kJ
Answer:
maintain
Explanation:
i don't know it's a guess
Answer:
The tension in the string is 78.73 N.
Explanation:
The tension in the string can be determined from the expression;
v = 
where: v is the speed of the wave in the sting, T is the tension in the string and m is the mass per unit length of the sting.
Given that: v = 16.2 m/s, and m = 0.3 kg/m.
Then;
16.2 = 
Square both sides to have,
= 
T =
x 0.3
= 252.44 x 0.3
= 78.732
T = 78.732 N
The tension in the string is 78.73 N.