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Stells [14]
4 years ago
7

When making calculations, you should rely on the precision of your measured data.

Chemistry
1 answer:
Masteriza [31]4 years ago
7 0
True, because measured data is precise.
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What is the definition of a Lewis acid?
Sveta_85 [38]

Answer:

A Lewis acid is a chemical species that contains an empty orbital which is capable of accepting an electron pair from a Lewis base to form a Lewis adduct

Explanation:

CAN YOU MAKE ME BRAINELIST PLEASE

8 0
3 years ago
Which descriptions apply to emitted radiation? check all that apply?
Murrr4er [49]
<span>measurement in Ci/Bq

the amount of radioactive materials released into the environment.

number of disintegrations of radioactive atoms in a radioactive material over a period of time</span>
7 0
3 years ago
A gas mixture at 535.0°C and 109 kPa absolute enters a heat exchanger at a rate of 67.0 m3/hr. The gas leaves the heat exchanger
SVEN [57.7K]

Answer:

the heat rate required to cool down the gas from 535°C until 215°C is -2.5 kW.

Explanation:

assuming ideal gas behaviour:

PV=nRT

therefore

P= 109 Kpa= 1.07575 atm

V= 67 m3/hr = 18.6111 L/s

T= 215 °C = 488 K

R = 0.082 atm L /mol K

n = PV/RT = 109 Kpa = 1.07575 atm * 18.611 L/s /(0.082 atm L/mol K * 488 K)

n= 0.5 mol/s

since the changes in kinetic and potencial energy are negligible, the heat required is equal to the enthalpy change of the gas:

Q= n* Δh = 0.5 mol/s * (- 5 kJ/mol) =2.5 kW

7 0
3 years ago
The total number of atoms in the formula aluminum acetate is
REY [17]

Answer:

B) 22 atoms

Explanation:

Aluminum Acetate has a formula of C6H9AlO6, meaning that it has 6 Carbon atoms, 9 Hydrogen atoms, 1 Aluminum atom, and 6 Oxygen atoms.

In total there are: 6+9+1+6 atoms, or a total of 22 atoms.

Hope this helps!

7 0
3 years ago
Read 2 more answers
The half-life for the radioactive decay of ce−141 is 32.5 days. if a sample has an activity of 3.8 μci after 162.5 d have elapse
Umnica [9.8K]
Answer : 121.5 <span>μCi

Explanation : We have Ce-141 half life given as 32.5 days so if the activity is 3.8 </span><span>μci after 162.5 days of time elapsed we have to find the initial activity.

We can use this formula;

</span>\frac{N}{ N_{0} } =  e^{-( \frac{0.693 X  T_{2} }{T_{1}})

3.8 / N_{0} = e^ ((0.693 X 162.5 ) / 32.5) = 121.5
<span>
On solving we get, The initial activity as 121.5  </span>μci
5 0
3 years ago
Read 2 more answers
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