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stiv31 [10]
3 years ago
11

The element with the greatest density is osmium, which has

Chemistry
1 answer:
Vinil7 [7]3 years ago
5 0

Answer:

Mass is 725.46g/cm

Explanation:

Mass = density × volume = 22.6g/cm^3 × 32.1cm^2 = 725.46g/cm

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a nuclide of 64/29 cu absorbs a positron. witch is the resulting atom? (A) 65/29Cu (B) 63/29 CU (C) 64/28Ni (D) 64/30 Zn
Nesterboy [21]

Answer : Option D) Zn^{64}_{30}

Explanation : When a positron is getting absorbed it means it will be e^{+1}_{0} so, the Cu^{64}_{29} will get converted;

So, the whole reaction will be;

Cu^{64}_{29} + e^{+1}_{0} ----> Zn^{64}_{30}.

This will convert the whole element of Cu will get changed into Zn. As, it absorbs by the positron, the atomic number gets increased from 29 to 30.

4 0
3 years ago
What behavior of light PROVES that it is a wave? Hint: will this ONLY happen because light is a wave, or could it also be true o
irina [24]

Answer:

Light changes speed in a glass of water.

Explanation:

8 0
3 years ago
If the concentration of Mg2+ in the solution were 0.039 M, what minimum [OH−] triggers precipitation of the Mg2+ ion? (Ksp=2.06×
Elodia [21]

Answer:

2.30 × 10⁻⁶ M

Explanation:

Step 1: Given data

Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M

Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³

Step 2: Write the reaction for the solution of Mg(OH)₂

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)

Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂

We will use the following expression.

Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²

[OH⁻] = 2.30 × 10⁻⁶ M

3 0
3 years ago
3. What noble gas would be part of the electron configuration notation for Mn?
shusha [124]

Answer:

Argon {Ar}

Explanation:

The noble gas used for a condensed electron configuration is the one before the element which you are configuring.

In this case, the element (Mn) is manganese

The noble gas that is before this element is Argon which is the row above it

so your configuration would be {Ar} 3d^5 4s^2

7 0
3 years ago
Write the balanced equation. Then calculate the volume of 0.65 M HCl required to completely neutralize 400.0 ml of 0.88 M KOH.
Misha Larkins [42]
Hello!

The balanced equation for the neutralization of KOH is the following:

HCl(aq) + KOH(aq) → KCl(aq) + H₂O(l)

To calculate the volume of HCl required, we can apply the following equation:

 moles HCl = moles KOH \\  \\ cHCl*vHCl=cKOH*vKOH \\  \\ vHCl= \frac{cKOH*vKOH}{cHCl}= \frac{400 mL*0,88M}{0,65M}=  541,54mL

So, the required volume of HCl is 541,54 mL

Have a nice day!
8 0
3 years ago
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