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NNADVOKAT [17]
3 years ago
15

How would it have been if particles of matter had no space?

Chemistry
1 answer:
iragen [17]3 years ago
6 0

Answer:

Matter is made of small particles of atoms or molecules. ... The particles in the solid are touching with very little space between them. The particles in a liquid usually are still touching but there are some spaces between them. The gas particles have big distances between them.

You might be interested in
How many moles are there in 67.2 grams of h2
yarga [219]

Answer: 33.3 moles

Explanation: 67.2 g H2 = 67.2/2.016 = 33.3 moles

7 0
3 years ago
A solution contains 0.036 M Cu2+ and 0.044 M Fe2+. A solution containing sulfide ions is added to selectively precipitate one of
Ratling [72]

Answer:

The precipitate is CuS.

Sulfide will precipitate at  [S2-]= 3.61*10^-35 M

Explanation:

<u>Step 1: </u>Data given

The solution contains 0.036 M Cu2+ and 0.044 M Fe2+

Ksp (CuS) = 1.3 × 10-36

Ksp (FeS) = 6.3 × 10-18

Step 2:  Calculate precipitate

CuS → Cu^2+ + S^2-         Ksp= 1.3*10^-36

FeS → Fe^2+ + S^2-      Ksp= 6.3*10^-18

Calculate the minimum of amount needed to form precipitates:

Q=Ksp

<u>For copper</u>  we have:  Ksp=[Cu2+]*[S2-]

Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]

[S2-]= 3.61*10^-35 M

<u>For Iron</u>  we have: Ksp=[Fe2+]*[S2-]

Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]

[S2-]= 1.43*10^-16 M

CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.

The precipitate is CuS.

Sulfide will precipitate at  [S2-]= 3.61*10^-35 M

3 0
3 years ago
Write a balanced half-reaction for the oxidation of aqueous hydrazine N2H4 to gaseous nitrogen N2 in basic aqueous solution
Zigmanuir [339]

Answer:

N2H2(aq) + 2OH^-(aq) ----------> N2(g) + 2H2O(l) + 2e

Explanation:

Hydrazine is mostly used in thermal engineering as an anticorrosive agent. Hydrazine can be oxidized in aqueous solution as shown in the equation above. Oxidation has to do with loss of electrons and increase in oxidation number.

The oxidation number of nitrogen in the equation increased from -1 in hydrazine on the lefthand side of the reaction equation to zero in nitrogen on the right hand side of the reaction equation. Two electrons were lost in the process as shown.

7 0
3 years ago
When Na and Cl combine to form a compound, which atom(s) will become a cation?
vivado [14]
Na only because cation is a positive ion whole cl is a negative ion in anion
8 0
3 years ago
Read 2 more answers
When electromagnetic radiation of wavelength 300nm falls on the surface of sodium electrons are emitted with a KE of 1.68 * 10 5
gtnhenbr [62]

Answer:

3.83 × 10⁻¹⁹ J;  518 nm  

Step-by-step explanation:

The equation for the <em>photoelectric effect</em> is

hf = Φ + KE  

<em>Data: </em>

λ = 300 nm = 300 × 10⁻⁹ m

KE = 1.68 × 10⁵ J/mol

Calculations:

Part 1. Minimum energy to remove an electron

(a) Calculate the <em>energy of the photon</em>

fλ = c  

 f = c/λ     Divide each side by λ

E = hf

E = hc/λ

E = (6.626× 10⁻³⁴ × 2.998 × 10⁸)/(300 × 10⁻⁹)

E = 6.622 × 10⁻¹⁹ J

(b) Calculate the <em>KE of one electron</em>

KE = 1.68 × 10⁵ × 1/(6.022 × 10²³)

KE = 2.790 × 10⁻¹⁹ J

(c) Calculate the work function

hf = Φ + KE     Subtract KE from each side

Φ = 6.622 × 10⁻¹⁹  - 2.790 × 10⁻¹⁹

Φ = 3.83 × 10⁻¹⁹ J

The minimum energy to remove an electron from a sodium atom

is 3.83 × 10⁻¹⁹ J.

Part 2. Maximum wavelength to remove an electron

The photon must have just enough energy to overcome the work function and leave the electron with zero kinetic energy.

    E = Φ

hc/λ = Φ                      Multiply each side by λ

  hc = Φ λ                   Divide each side by Φ

   λ = hc/ Φ

   λ = (6.626 × 10⁻³⁴ × 2.998 × 10⁸)/(3.83 × 10⁻¹⁹)

   λ = 5.18 × 10⁻⁷ m     Convert to nanometres

   λ = 518 nm

The maximum wavelength that will cause an electron to move is 518 nm.

6 0
3 years ago
Read 2 more answers
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