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Katena32 [7]
2 years ago
15

What volume is equivalent to 0.0015 m3?

Chemistry
1 answer:
vaieri [72.5K]2 years ago
3 0

Answer : Option 'D' is correct.

0.0015m^{3} = 1.5\times 10^{6}mm^{3}

Explanation 1) : 1 m = 100 cm

1 m^{3}= 1000000 cm^{3} or 1 m^{3}= 10^{6} cm^{3}

(0.0015m^{3})\times \frac{(1000000cm^{3})}{(1m^{3})}

(0.0015)\times \frac{(1000000cm^{3})}{(1)} = 1500cm^{3}

0.0015m^{3} = 1500cm^{3}  = 1.5\times 10^{3}cm^{3}

Explanation 2) : 1 m = 1000 mm

1m^{3}= 1000000000mm^{3} = 10^{9}mm^{3}

(0.0015m^{3})\times \frac{(10^{9}mm^{3})}{(1m^{3})}

(0.0015)\times \frac{(10^{9}mm^{3})}{(1)} = 1500000 mm^{3}

0.0015m^{3} = 1500000 mm^{3} = 1.5\times 10^{6}mm^{3}

Option 'D' is right.


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B

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7 0
2 years ago
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Write the balanced Ka and Kb reactions for HSO3- in water. Be sure to include the physical states of each species involved in th
Hunter-Best [27]

Answer:

Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]

Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]

Explanation:

An amphoteric substance as HSO₃⁻ is a substance that act as either an acid or a base. When acid:

HSO₃⁻(aq) + H₂O(l) ⇄ H₃O⁺(aq) + SO₃²⁻(aq)

And Ka, the acid dissociation constant is:

<h3>Ka = [H₃O⁺] [SO₃²⁻] / [HSO₃⁻]</h3><h3 />

When base:

HSO₃⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + H₂SO₃(aq)

And kb, base dissociation constant is:

<h3>Kb = [OH⁻] [H₂SO₃] / [HSO₃⁻]</h3>

6 0
1 year ago
A large flask is evacuated and weighed, filled with argon gas, and then reweighed. When reweighed, the flask is found to have ga
Elden [556K]

Answer:

100.52

Explanation:

from the ideal gas equation PV=nRT

for a given container filled with any ideal gas P and V remains constant.So T is also constant.R is as such a constant.

So n i.e no of moles will also be constant.

no of moles of Ar=3.224/40=0.0806

no of moles of unknown gas=0.0806

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2 years ago
Yousef measured the height of each seedling on day 1 and day 7. These are his results.
kati45 [8]
That’s a lot of numbers
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2 years ago
A 0.133 mol sample of gas in a 525 ml container has a pressure of 312 torr. The temperature of the gas is ________ °c.
Serggg [28]

Answer:

-253.2 ^{\circ}C

Explanation:

First of all, we need to convert the pressure of the gas from torr to Pa. We know that:

1 torr = 133.3 Pa

So, the pressure in Pascals is

p=(312 torr)(133.3 Pa/torr)=4.16\cdot 10^4 Pa

Then we also have:

n = 0.133 number of moles of the gas

V=525 mL=0.525 L=5.25\cdot 10^{-4} m^3 volume of the gas

The ideal gas equation states that

pV=nRT

where R is the gas constant and T the absolute temperature. Solving the equation for T, we find

T=\frac{pV}{nR}=\frac{(4.16\cdot 10^4 Pa)(5.25\cdot 10^{-4} m^3)}{(0.133 mol)(8.314 J/mol K)}=19.8 K

In Celsius, it becomes

T=19.8 K-273=-253.2 ^{\circ}C

3 0
2 years ago
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