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2 years ago

What is the chemical name for K2Se

Chemistry

2 answers:

german2 years ago

8 0

Answer:

Potassium selenide

Explanation:

Oduvanchick [21]2 years ago

6 0

Answer:

Potassium selenide

Explanation:

Potassium selenide (K2Se)

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In the gaseous state, chlorine exists as a diatomic molecule Cl2 (Molar mass = 70.9 g/mol). Calculate the number of moles of chl

Annette [7]

Answer:

3.67 mol Cl

Explanation:

We need to convert g of Cl 2 to moles of Cl. First we divide 130 gCl2 by the molar mass (70.90 gCl2/mol) to find out how many moles of Cl2 do we have.

130 gCl2 x $\frac{1 mol Cl2 }{70.90 gCl2}$ = 1.83 mol Cl2

Then we need to convert 1.83 mol de Cl2 to moles of Cl. We have 2 moles of Cl in every Cl2 molecule so we just need to multiply by 2.

1.83 molCl2 x $\frac{2 molCl}{1 molCl2}$ = 3.67 molCl

8 0

3 years ago

Read 2 more answers

Hydrogen sulfide burns form sulfur dioxide:

Helga [31]

Answer: 404.04 kJ.

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

moles of $H_2S$

$\text{Number of moles}=\frac{26.7g}{34.1g/mol}=0.78moles$

$2H_2S(g)+3O_2(g)\rightarrow 2SO_2(g)+2H_2O(g)$ $\Delta H=-1036kJ$

According to stoichiometry :

2 moles of $H_2S$ on burning produces = 1036 kJ

Thus 0.78 moles of $H_2S$ on burning produces = $\frac{1036kJ}{2}\times 0.78=404.04$

Thus the enthalpy change when burning 26.7 g of hydrogen sulfide is 404.04 kJ.

7 0

3 years ago

Which table correctly identifies the subatomic particle's charge and mass?<br> a<br> b<br> c<br> d

s344n2d4d5 [400]

It’s b that charts say it

8 0

3 years ago

A nuclear power plant operates at 40.0% efficiency with a continuous production of 1042 MW of usable power in 1.00 year and cons

Sergio039 [100]

Answer:

3.00 x 10^-11 joules / atom of U-235

Explanation:

We know that the formula for Power = Work done (w)/Time (t)

We need to get the joules from power , since Joules is the SI unit of work.

From the formula P = W/t

W = Power (P) * Time (t)

The SI unit for Time is seconds, hence we change 1 year in seconds

1yr * 365 days/yr * 24hrs/day * 60mins/hr * 60 secs/min = 31536000 secs

It was stated in the question that the plant operates at an efficiency of 40%,

Thus to get the true power we divide the power provided in the question by 0.4 or 40%

= X(0.4) = 1042MW

True Power X = 1042/0.4 = 2605MW

Thus true power = 2605 * 10^6 Watts

Now we have the time in seconds and true power in Watts, we then find the work done.

From our above formula P = W/t

W = P*t = (2605* 10^6) (31536000) =

Finally, we can solve for our energy (work):

P = W / T PT = W = (2880x10^6) (31536000) = 8.22 x 10^16 joules

We then calculate the amount of energy released by only 1 single uranium-235 atom.

= 8.22 x 10^16 joules / 1.07x10^6 g U-235 (235 g / 1 mol)(1 mol/6.0210^23 atoms)

= 3.00 x 10^-11 joules / atom of U-235

5 0

3 years ago

Period, block, and group

Sedaia [141]

Answer:

what do you mean

Explanation:

8 0

3 years ago