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zubka84 [21]
3 years ago
6

If there is a New Moon at midnight on October 1, when would the next New Moon occur?

Chemistry
1 answer:
irina1246 [14]3 years ago
8 0
Answer: C

Explanation: A lunar month is 29.53 earth days. Since an earth calendar as only integer days, it could be either 29 or 30 days later.
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Drag the tiles to the correct locations on the equation. Not all tiles will be used.
mel-nik [20]

Answer:

Tile 52 and three are incorrect

Explanation:

6 0
2 years ago
A chemistry graduate student is given of a chlorous acid solution. Chlorous acid is a weak acid with . What mass of should the s
DerKrebs [107]

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>
5 0
3 years ago
What is the purpose of finding oxidation states in the half-reaction method for balancing equations?
vazorg [7]

Answer:

B. To Identify the half reactions for the equation

4 0
3 years ago
In a Dam, If we doubled the depth of the dam the hydrostatic force will be ?
AnnZ [28]

Answer: Option (A) is the correct answer.

Explanation:

Force acting on a dam is as follows.

                  F = \frac{1}{2}\rho g\omega H^{2} .......... (1)

Now, when we double the depth then it means H is increasing 2 times and then the above relation will be as follows.

                F' = \frac{1}{2}\rho g\omega (2)^{2}

               F' = \frac{1}{2}\rho g\omega 4 ........... (2)

Now, dividing equation (1) by equation (2) as follows.

          \frac{F}{F'} = \frac{\frac{1}{2}\rho g\omega H^{2}}{\frac{1}{2}\rho g\omega 4}  

Cancelling the common terms we get the following.

                 \frac{F}{F'} = \frac{1}{4}    

                   4F = F'

Thus, we can conclude that if doubled the depth of the dam the hydrostatic force will be 4F.

4 0
3 years ago
Would precipitation occur when 500 mL of a 0.02M solution of AgNO3 is mixed with 500 mL of a 0.001M solution of NaCl? Show your
Oksanka [162]
We know,
AgNO3 + NaCl ⇒ NaNO3 + AgCl(s)
The moles of Na+ present:
0.5 L * 0.001 mol/L
= 5 x 10⁻⁴ mol
Moles of Ag+ present:
0.5 * 0.02
= 0.01 mol
The limiting reactant is Na
Therefore, the moles of Ag reacted:
5 x 10⁻⁴
AgCl is insoluble in water; therefore, the AgCl formed will precipitate
7 0
3 years ago
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