Magma that cools quickly forms one kind of igneous rock, and magma that cools slowly forms another kind. When magma rise from deep within the earth and explodes out of a volcano, it is called lava, and it cools quickly on the surface. Rock formed in this way is called extrusive igneous rock.
Kepler's
third law shows the relationship between the orbital period of an object and
the distance between the object and the object it orbits.
The
simplified version of this law is: P^2 = a^3
Where,
P =
period of the orbit in years = 0.62 years
a =
average distance from the object to the object it orbits in AU. The
astronomical unit AU is a unit of length which is roughly equivalent to the
distance from Earth to the Sun.
Therefore
calculating for a:
0.62
^ 2 = a ^ 3
a =
0.62 ^ (2/3)
a =
0.727 AU = 0.72 AU
Therefore we can interpret this as: The distance from Venus to the Sun is about 72% of the distance from Earth to
Sun.
<span>Answer:
B. 0.72 AU</span>
Answer:
11.31g NaClO₂
Explanation:
<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>
It is possible to answer this question using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>
You can change the concentration of the substance if you write the moles of the substances:
[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>
Replacing in H-H expression, as the pH you want is 1.45:
1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]
0.1251 = Moles NaClO₂
As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:
0.1251 moles NaClO₂ ₓ (90.44g / mol) =
<h3>11.31g NaClO₂</h3>
Answer:
The scientific laws have been well proven before they are published so it is difficult to prove mistakes
Explanation:
Answer:
V KOH = 41 mL
Explanation:
for neutralization:
- ( V×<em>C </em>)acid = ( V×<em>C </em>)base
∴ <em>C </em>H2SO4 = 0.0050 M = 0.0050 mol/L
∴ V H2SO4 = 41 mL = 0.041 L
∴ <em>C</em> KOH = 0.0050 N = 0.0050 eq-g/L
∴ E KOH = 1 eq-g/mol
⇒ <em>C</em> KOH = (0.0050 eq-g/L)×(mol KOH/1 eq-g) = 0.0050 mol/L
⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH
⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)
⇒ V KOH = 0.041 L