All categories

2 years ago

What volume is equivalent to 0.0015 m3?

Chemistry

1 answer:

Natali [406]2 years ago

8 0

1.5 x 10⁶mm³

Explanation:

Here we are to convert m³ to mm³ or cm³ to see the correct option.

100cm = 1 m

1000mm = 1m

Converting to cm³ from m³;

0.0015m³ = 1.5 x 10⁻³m³

Now:

1.5 x 10⁻³ x m x m x m x $\frac{100cm}{1m}$ x $\frac{100cm}{1m}$ x $\frac{100cm}{1m}$

= 1.5 x 10³cm³

Also;

1.5 x 10⁻³ x m x m x m x $\frac{1000mm}{1m}$ x $\frac{1000mm}{1m}$ x $\frac{1000mm}{1m}$

= 1.5 x 10⁶mm³

learn more:

Volume brainly.com/question/2690299

#learnwithBrainly

You might be interested in

The specific rotation of (R) carvone is (+) 61°. The optical rotation of a sample of a mixture of R &S carvone is measured a

shusha [124]

Answer:

See explanation

Explanation:

% optical purity = specific rotation of mixture/specific rotation of pure enantiomer * 100/1

specific rotation of mixture = 23°

specific rotation of pure enantiomer = 61°

Hence;

% optical purity = 23/61 * 100 = 38 %

More abundant enantiomer = 100% - 38 % = 62%

Hence the pure (S) carvone is (-) 62° is the more abundant enantiomer.

Enantiomeric excess = 62 - 50/50 * 100 = 24%

Hence

(R) - carvone = 38 %

(S) - carvone = 62%

7 0

1 year ago

Read 2 more answers

Combined gas law problem: a balloon is filled with 500.0 mL of helium at a temperature of 27 degrees Celsius and 755 mmHg. As th

Papessa [141]

The statement of the combined gas law for a fixed amount of gas is,
PV/T = constant
Here, the units of pressure and volume must be consistent and the temperature must be the absolute temperature (Kelvin or Rankine).
0.65 atm is equivalent to 494 mmHg
Using the equation:
(755 x 500) / (27 + 273) = (494 x V) / (-33 + 273)
V = 3396 ml = 3.4 liters

8 0

2 years ago

Write the net ionic equation for the reaction of zinc metal with aqueous iron(II) nitrate. Include physical states.

kotykmax [81]

Answer:

Zn°(s) + Fe⁺²(aq) => Zn⁺²(aq) + Fe°(s)

Explanation:

Molecular Equation:

Zn°(s) + Fe(NO₃)₂(aq) => Zn(NO₃)₂(aq) + Fe°(s)

Ionic Equation:

Zn°(s) + Fe⁺²(aq) + 2NO₃⁻(aq) => Zn⁺²(aq) + 2NO₃⁻(aq) + Fe°(s)

Net Ionic Equation: => Drop NO₃⁻ as spectator ion

Zn°(s) + Fe⁺²(aq) => Zn⁺²(aq) + Fe°(s)

7 0

2 years ago

At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2 + O2(g) → 2SO3(g) At equilibrium, the partia

lapo4ka [179]

Answer:

The partial pressure of SO₃ is 82.0 atm

Explanation:

The equilibrium constant Kp is equal to <em>the equilibrium pressure of the gaseous products raised to the power of their stoichiometric coefficients divided by the equilibrium pressure of the gaseous reactants raised to the power of their stoichiometric coefficients</em>.

For the reaction,

2 SO₂(g) + O₂(g) → 2 SO₃(g)

$Kp = 0.345 = \frac{(pSO_{3})^{2} }{(pSO_{2})^{2} \times pO_{2} }\\pSO_{3} = \sqrt[]{0.345 \times (pSO_{2})^{2} \times pO_{2} } \\pSO_{3} = \sqrt[]{0.345 \times (35.0)^{2} \times 15.9 } \\pSO_{3} = 82.0 atm$