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Mazyrski [523]
3 years ago
15

Which type of radioactive decay does not produce any particles?

Chemistry
1 answer:
natta225 [31]3 years ago
3 0
The answer is is gamma as alpha an beta both result in the release of a particle

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6) A student measures out 96.21 g of sulfur for an experiment. How many moles of Sulfur are in this
barxatty [35]

Atomic mass of Sulfur = 32g

32g of Sulfur is one mole.

1g of Sulfur is \frac{1}{32} moles

96.21g of Sulfur is \frac{96.21}{32} moles=> 3moles(appx)

5 0
3 years ago
Read 2 more answers
3.<br> КОН<br> +<br> —<br> —<br> Н3РО4<br> 1<br> +<br> K3PO4<br> —<br> Н,0<br> —
expeople1 [14]
What does this mean?
3 0
2 years ago
Which of the following is transferred due to a temperature difference?
quester [9]
Water can be turned to ice if to cold gas if to hot 
ice can turn to water if to hot and stay the same if to cold 
gas will turn to water if to hot and freeze to ice if to cold and the pattern keeps going like that.
hope this helps 
 
8 0
2 years ago
When adjusted for any changes in ΔHΔH and ΔSΔS with temperature, the standard free energy change ΔG∘TΔGT∘Delta G_{T}^{\circ} at
STALIN [3.7K]

The equilibrium constant is 0.0022.

Explanation:

The values given in the problem is

ΔG° = 1.22 ×10⁵ J/mol

T = 2400 K.

R = 8.314 J mol⁻¹ K⁻¹

The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.

The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K

So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.

We get,

1.22 * 10^{5} = - 8.314* 2400 * ln K

\\ 1.22 * 10^{5} = -19953.6 * ln K

ln K = \frac{-1.22*10^{5} }{19953.6} =-6.114\\\\k =e^{-6.114}=0.0022

So, the equilibrium constant is 0.0022.

4 0
3 years ago
In a coffee-cup calorimeter, 1.55 g KOH is added to 125 ml of 0.80 M HCl. The following reaction occurs. KOH(s) + HCl(aq) → H2O(
DENIUS [597]

Answe1.55 gr:

Explanation:

yes

4 0
3 years ago
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