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Dima020 [189]
3 years ago
5

How many grams of NaCl are needed to make 1.00 liter of a 3.00 M NaCl solution?

Chemistry
1 answer:
Mkey [24]3 years ago
4 0
Do they have a Grapher anything
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How do these ion things work?
castortr0y [4]

Answer:

ion know figure it out its you who learning ab it

Explanation:

5 0
3 years ago
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When aqueous solutions of chromium(II) iodide and sodium phosphate are combined, solid chromium(II) phosphate and a solution of
Georgia [21]

Answer:

3 Cr²⁺(aq) + 2 PO₄³⁻(aq) ⇄ Cr₃(PO₄)₂(s)

Explanation:

When aqueous solutions of chromium(II) iodide and sodium phosphate are combined, solid chromium(II) phosphate and a solution of sodium iodide are formed. The molecular equation is:

3 CrI₂(aq) + 2 Na₃PO₄(aq) ⇄ Cr₃(PO₄)₂(s) + 6 NaI(aq)

The full ionic equation includes all the ions and the molecular species.

3 Cr²⁺(aq) + 6 I⁻(aq) + 6 Na⁺(aq) + 2 PO₄³⁻(aq) ⇄ Cr₃(PO₄)₂(s) + 6 Na⁺(aq) +         6 I⁻(aq)

The net ionic equation includes only the ions that participate in the reaction and the molecular species.

3 Cr²⁺(aq) + 2 PO₄³⁻(aq) ⇄ Cr₃(PO₄)₂(s)

4 0
4 years ago
What are two sources of phosphorous?
aleksandr82 [10.1K]
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3 years ago
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Lithium ions in Lithium selenide (Li2Se) have an atomic radius of 73 pm whereas the selenium ion is 184 pm. This compound is mos
NARA [144]

Explanation:

Formula according to the radius ratio rule is as follows.

             \frac{r_{+}}{r_{-}} = \frac{73}{184}

                          = 0.397

According to the radius ratio rule, as the calculated value is 0.397 and it lies in between 0.225 to 0.414. Therefore, it means that the type of void is tetrahedral.

Thus, we can conclude that the given compound is most likely to adopt closest-packed array with lithium ions occupying tetrahedral holes.

3 0
3 years ago
Determine the frequency and wavelength (in nm) of the light obsorded when the e- goes from n=1 to n=5
morpeh [17]

Answer:

Frequency = 3.16 ×10¹⁴ Hz

λ = 0.95×10² nm

Explanation:

Energy associated with nth state is,

En =  -13.6/n²

For n = 1

E₁ = -13.6 / 1²

E₁ = -13.6/1

E₁ = -13.6 ev

Kinetic energy of electron = -E₁ = 13.6 ev

For n = 5

E₅ = -13.6 / 5²

E₅ = -13.6/25

E₅ = -0.544 ev

Kinetic energy of electron = -E₅ = 0.544 ev

Wavelength of radiation emitted:

E = hc/λ = E₅ - E₁

hc/λ = E₅ - E₁

by putting values,

6.63×10⁻³⁴Js × 3×10⁸m/s / λ = -0.544ev  - (-13.6 ev  )

6.63×10⁻³⁴ Js× 3×10⁸m/s / λ = 13.056 ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /13.056 ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /13.056 × 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 13.056 × 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 20.9 ×10⁻¹⁹ J

λ = 0.95×10⁻⁷ m

m to nm:

0.95×10⁻⁷ m ×10⁹nm/1 m

0.95×10² nm

Frequency:

Frequency = speed of electron / wavelength

by putting values,

Frequency = 3×10⁸m/s /0.95×10⁻⁷ m

Frequency = 3.16 ×10¹⁴ s⁻¹

s⁻¹ = Hz

Frequency = 3.16 ×10¹⁴ Hz

4 0
3 years ago
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