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Phoenix [80]
3 years ago
6

Susana heats up a sample of red crystals. While the sample is being heated,a gas is released and a blue powder is left after hea

ting. Is the sample of red crystals an element or compound?
Chemistry
2 answers:
valkas [14]3 years ago
4 0

Answer:

<u>Susana's red crystal sample is a compound because it was broken down into a gas and blue powder. It is not an element because elements cannot be broken down into simpler substances by ordinary means, such as heating.</u>

Explanation

Sample response form Edg

harina [27]3 years ago
3 0

Answer & Explanation:

Susana's red crystal sample is a compound because it was broken down into a gas and blue powder. It is not an element because elements cannot be broken down into simpler substances by ordinary means, such as heating.

You might be interested in
How much heat energy, in kilojoules, is required to convert 53.0 g of ice at −18.0 ∘C to water at 25.0 ∘C ? Express your answer
ANTONII [103]

Answer:

25.2 kJ of heat energy is required to convert 53.0 g of ice at −18.0 ∘C to water at 25.0 ∘C

Explanation:

Data: mass of ice = m = 53.0 g

         Temperature of ice = T1 = -18.0 ∘C

         Temperature of water = T2 = 25.0 ∘C

          Change in Temperature = ΔT : T2-T1                                          

           Specific heat of ice = c(i) = 2.09 J/g . ∘C

          Specific heat of water = c(w) = 4.18 J/g . ∘C

          Enthalpy of fusion of water (when convert from ice to water) = ΔH(f) : 334 J/g

          Enthalpy of vapourization of water (when convert from water to gas) = ΔH(v) :2250 J/g

          Total heat required = q = ?

Solution:        Heat required to melt the ice

                      T1 = -18 ∘C

                      T2 = 0 ∘C

                      Q1 = m x c(i) x ΔT

                      Q1 = 53.0 x 2.09 x (0-(-18))

                      Q1 = 53.0 x 2.09 x 18

                      Q1 = 1993.86 J is required by ice to reach to its

                      melting point.

                      <u>Heat required to convert the ice into water</u>

                      Q2 = ΔH(f) x m

                      Q2 = 334 x 53.0 = 17702 J is required by ice to  

                      convert into water.

                      <u> Heat required by water to reach at 25 ∘C</u>

                       T1 = 0 ∘C

                       T2 = 25 ∘C

                       Q3 = m x c(w) x ΔT

                       Q3 = 53.0 x 4.18 x (25-0)

                       Q3 = 5538.5 J is required by water to reach at  

                       25∘C .

                       <u>Total heat required by 53.0 g of ice at −18.0 ∘C  </u>

                        <u>to water at </u>

                        <u>25.0 ∘C</u>

                        q = Q1 + Q2 + Q3

                        q = 1993.86 J + 17702 J + 5538.5 J

                        q = 25234.36 J is the total heat required 53.0 g  

                        of ice at −18.0 ∘C to water at 25.0 ∘C.

                        <u>Conversion from Joule (J) to Kilojoule (kJ)</u>

                          1000 J =  1 kJ

                          q = 25234.36 J/1000

                          q = 25.23436 kJ

                           <u>Conversion of heat in kJ to three significant   </u>

                           <u>figures</u>

                           q = 25.2 kJ

8 0
4 years ago
Consider the data presented below. time (s) 0 40 80 120 160 moles of a 0.100 0.067 0.045 0.030 0.020 part a part complete determ
Whitepunk [10]
To determine which order of the reaction it is, first we need to calculate the rate of change of moles.
the data is as follows 
time         0         40        80       120       160
moles    0.100   0.067  0.045    0.030    0.020


Q1)
for the first 40 s change of moles ;
      = -d[A] / t
      = - (0.067-0.100)/40s
      = 8.25 x 10⁻⁴ mol/s
for the next 40 s
      = -(0.045-0.067)/40
      = 5.5 x 10⁻⁴ mol/s
the 40 s after that
      = -(0.030-0.045)/40 s 
     = 3.75 x 10⁻⁴ mol/s
k - rate constant
and A is the only reactant that affects the rate of the reaction

rate = k [A]ᵇ
8.25 × 10⁻⁴ mol/s = k [0.100 mol]ᵇ ----1
5.5 x 10⁻⁴ mol/s = k [0.067 mol]ᵇ   -----2
divide the 2nd equation by the 1st equation
1.5 = [1.49]ᵇ
b is almost equal to 1
Therefore this is a first order reaction

Q2)
to find out the rate constant(k), we have to first state the equation for a first order reaction.
rate = k[A]ᵇ
As A is the only reactant thats considered for the rate equation. 
Since this is a first order reaction,
b = 1
therefore the reaction is 
rate = k[A]
substituting the values,
8.25 x 10⁻⁴ mol/s = k [0.100 mol]
k = 8.25 x 10⁻⁴ mol/s /0.100mol
   = 8.25 x 10⁻³ s⁻¹

7 0
4 years ago
If an element forms 1- ion, in which group of the periodic table would you expect to find it
irina1246 [14]

Many of the elements on the periodic table will always form ions that have the same charge. The alkali metals (shown in yellow) always form +1 ions. The alkaline earth metals (red) always form +2 ions. The halogens (blue) always form -1 ions.

4 0
3 years ago
Read 2 more answers
What change will be caused by addition of a small amount of hcl to a solution containing fluoride ions and hydrogen fluoride?
agasfer [191]

Answer:

  • <em>The addition of a small amount of HCl to a solution containing fluoride ions and hydrogen fluoride</em> <u>will cause the equilibrium shift to the production of more hydrogen fluoride.</u>

Explanation:

The solution containing <em>fluoride ions and hydrogen fluoride</em> (a weak acid) may be chemically represented by this equilibrium equation:

  • F⁻ (aq) + H⁺ (aq) ⇄ HF (aq)

The <em>HCl</em>, a strong acid, added will ionize in water according to this chemical equation:

  • HCl (aq)  → H⁺ (aq) + Cl⁻ (aq)

Then, following Le Chatelir's principle, the addtion of H⁺ ions coming from the HCl dissociation, will increase the concentration of H⁺ in the solution, driving to the consumption of some F⁻ ions, and the production of more HF acid. This is a shift of the equilbrium toward the HF side.

8 0
3 years ago
Consider the following unbalanced redox reactions. In each case, separate the whole reactions into half-reactions, balance the h
Marina86 [1]

Answer & Explanation:

(a)

Fe^{2+} +NO_{3}^{-}  => Fe{(OH)}_3 + N_2

reducing agent = Fe²⁺

Oxidizing agent = NO₃⁻

oxidation

Fe²⁺  ⇒ Fe(OH)₃

reduction

NO₃⁻  ⇒ N₂

Oxidation Half Reaction

(<em>redox reactions are balanced by adding appropriate H⁺ and H₂O atoms)</em>

Fe²⁺ ⇒ Fe(OH)₃

Balance O atoms

Fe²⁺ + 3H₂O ⇒ Fe(OH)₃

Balance H atoms

Fe² + 3H₂0 ⇒ Fe(OH)₃ + 3H⁺

balance Charge

Fe² + 3H₂0 ⇒ Fe(OH)₃ + 3H⁺ + e⁻..............(1)

reduction Half Reaction

NO₃⁻ ⇒ N₂

Balance N atoms

2NO₃⁻ ⇒N₂

Balance O atoms by adding appropriate H₂O

2NO₃⁻ ⇒ N₂ + 6H₂O

Balance H atoms

2NO₃⁻ + 12H⁺ ⇒ N₂ + 6H₂O

Balance Charge

2NO₃⁻ + 12H⁺ + 10e⁻⇒ N₂ + 6H₂O.................(2)

Combine Equation (1) and (2)

(1) × 10: 10Fe² + 30H₂0 ⇒ 10Fe(OH)₃ + 30H⁺ + 10e⁻

(2) × 1:   2NO₃⁻ + 12H⁺ + 10e⁻⇒ N₂ + 6H₂O

(1) + (2): 10Fe² + <u><em>30H₂0</em></u> + 2NO₃⁻ + <u><em>12H⁺</em></u> + <u><em>10e⁻</em></u> ⇒10Fe(OH)₃ + <u><em>30H⁺</em></u><u><em> </em></u>+ <em><u>10e⁻</u></em> +              

            N₂ + <u><em>6H₂O</em></u>

            10Fe² + 24H₂0 + 2NO₃⁻  ⇒ 10Fe(OH)₃ + 18H⁺ + N₂

this is the balanced reaction

REDUCTION POTENTIAL

10Fe²⁺(aq) + 10e⁻ ⇒ 10Fe(OH)₃(aq)             E°ox = 10(-0.44) = -4.4V

2NO₃⁻(aq) -  2e⁻ ⇌ N₂(g) + 18H⁺       E°red = 2(+0.80) = +1.6

10Fe² + 24H₂0 + 2NO₃⁻  ⇒ 10Fe(OH)₃ + 18H⁺ + N₂    E°cell = -2.8V

E°cell = E°red + E°ox

4 0
4 years ago
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