Select the correct answer.

Enter the appropriate symbol for an isotope of potassium-39 corresponding to the isotope notation A/Z x

Taya2010 [7]

Appropriate symbol for an isotope of potassium - 39 corresponding to the isotope notion r A ZX

7 0

3 years ago

To test the effectiveness of a gunpowder mixture, 1 gram was exploded under controlled STP conditions, and the reaction chamber

Alekssandra [29.7K]

Answer:

1.3 × 10³ cm³

Explanation:

The gas occupies a volume of V₁ = 310 cm³ under standard temperature and pressure (STP), that is, T₁ = 273.15 K and P₁ = 1.0 atm. In order to find the volume V₂ under different conditions we can use the combined gas law formula.

$\frac{P_{1}.V_{1}}{T_{1}} =\frac{P_{2}.V_{2}}{T_{2}} \\V_{2}=\frac{P_{1}.V_{1}.T_{2}}{T_{1}.P_{2}}=\frac{1.0atm\times 310cm^{3} \times 2473K }{273.15K \times 2.1atm} =1.3 \times 10^{3} cm^{3}$

7 0

3 years ago

Pls answer fast chem questions

gregori [183]

I don't recognize this problem, please make sure the input is complete.

7 0

3 years ago

Water (10 kg/s) at 1 bar pressure and 50 C is pumped isothermally to 10 bar. What is the pump work? (Use the steam tables.) O -7

11Alexandr11 [23.1K]

Explanation:

For an isothermal process equation will be as follows.

W = nRT ln $\frac{P_{1}}{P_{2}}$

It is given that mass is 10 kg/s or 10,000 g/s (as 1 kg = 1000 g). So, calculate number of moles of water as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{10000 g/s}{18 g/mol}$

= 555.55 mol/s

= 556 mol/s (approx)

As T = $50^{o}C$ or (50 + 273.15) K = 323.15 K. Hence, putting the given values into the above formula as follows.

W = nRT ln[/tex]\frac{P_{1}}{P_{2}}[/tex]

= $556 mol/s \times 8.314 J/ K mol K \times 323.15 K \times ln\frac{1}{10}$

= $556 mol/s \times 8.314 J/ K mol K \times 323.15 K \times -2.303$

= -3440193.809 J/s

Negative sign shows work is done by the pump. Since, 1 J = 0.001 kJ. Therefore, converting the calculated value into kJ as follows.

$3440193.809 J/s \times \frac{0.001 kJ}{1 J}$

= 3440.193 kJ/s

= 3451 kJ/s (approx)

Thus, we can conclude that the pump work is 3451 kJ/s.

6 0

3 years ago

Find the number of moles of in 571,0 g of tetraphosphorous octoxide.

mars1129 [50]

Answer:

Think sorry if i did it wrong The molar mass that is the mass of 6.022 X 1023 atoms of helium is 4g. Hence 1 mole of He = 4 g of He = 6.022 X 1023atoms of He. 1 g of He = (6.022 X 1023) / 4 atoms of He. 52 g of He = (6.022 X 1023X 52 ) / 4 atoms of He.

For Helium,

Number of atoms=Number of moles ×6.022×1023

No of Moles = 6.022×10

2312.044×10

23 = 2.

Now, atomic mass of He=4

Weight = 2×4 = 8 g

The molar mass that is the mass of 6.022 X 1023 atoms of helium is 4g. Hence 1 mole of He = 4 g of He = 6.022 X 1023atoms of He. 1 g of He = (6.022 X 1023) / 4 atoms of He. 52 g of He = (6.022 X 1023X 52 ) / 4 atoms of He.

As we know, gram atoms = gram mole

here, no of moles = 23/46 =2 so no of gram atoms are 2.

Explanation:

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