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kogti [31]
2 years ago
7

Napisz wzory sumaryczne tlenków o podanych nazwach

Chemistry
1 answer:
Alja [10]2 years ago
6 0

Answer:

kingkongkabayo jahsgajavafhabavwh

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What reagent is <br>used to test for starch<br>​
Elodia [21]

Answer:

iodine

Explanation:

In the presence of starch, iodine turns a blue/black colour. It is possible to distinguish starch from glucose (and other carbohydrates) using this iodine solution test. For example, if iodine is added to a peeled potato then it will turn black. Benedict's reagent can be used to test for glucose.

8 0
3 years ago
What is the name for the compound formed between nitrogen and magnesium. (show your work)
astra-53 [7]

Answer:

the compound formed between nitrogen and magnesium is Magnesium Nitride

7 0
2 years ago
At 300.0 K and 0.987 atm pressure, what will be the volume of 2.30 mol of Ne?
Nadya [2.5K]

Answer:

V = 57.39 L

Explanation:

Given that,

Temperature, T = 300 K

Pressure, P = 0.987 atm

No. of moles of Ne, n = 2.30 mol

We need to find the volume of Ne. We know that, the ideal gas law is as follows :

PV = nRT

Where

P is pressure and R is gas constant

V=\dfrac{nRT}{P}\\\\V=\dfrac{2.3\times 0.0821\times 300}{0.987 }\\\\V=57.39\ L

So, the volume of the Ne is 57.39 L.

3 0
2 years ago
In the reaction shown above , the?
matrenka [14]

Answer:

I think option (d) is right answer

8 0
3 years ago
(2 pts) The solubility of InF3 is 4.0 x 10-2 g/100 mL. a) What is the Ksp? Include the chemical equation and Ksp expression. MW
Aleonysh [2.5K]

Answer:

a) Ksp = 7.9x10⁻¹⁰

b) Solubility is 6.31x10⁻⁶M

Explanation:

a) InF₃ in water produce:

InF₃ ⇄ In⁺³ + 3F⁻

And Ksp is defined as:

Ksp = [In⁺³] [F⁻]³

4.0x10⁻²g / 100mL of InF₃ are:

4.0x10⁻²g / 100mL ₓ (1mol / 172g) ₓ (100mL / 0.1L) = <em>2.3x10⁻³M  InF₃. </em>Thus:

[In⁺³] = 2.3x10⁻³M  InF₃ × (1 mol In⁺³ / mol InF₃) = 2.3x10⁻³M  In⁺³

[F⁻] = 2.3x10⁻³M  InF₃ × (3 mol F⁻ / mol InF₃) = 7.0x10⁻³M F⁻

Replacing these values in Ksp formula:

Ksp = [2.3x10⁻³M  In⁺³] × [7.0x10⁻³M F⁻]³ = <em>7.9x10⁻¹⁰</em>

<em></em>

b) 0.05 moles of F⁻ produce solubility of InF₃ decrease to:

7.9x10⁻¹⁰ = [x] [0.05 + 3x]³

Where x are moles of In⁺³ produced from solid InF₃ and 3x are moles of F⁻ produced from the same source. That means x is solubility in mol / L

Solving from x:

x = -0.018 → False solution, there is no negative concentrations.

x = 6.31x10⁻⁶M → Right answer.

Thus, <em>solubility is 6.31x10⁻⁶M</em>

3 0
3 years ago
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