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uysha [10]
2 years ago
15

D. potassium + iodine →

Chemistry
1 answer:
aliina [53]2 years ago
5 0

Answer:

The missing reagents are.

Potassium + Iodine =<u> Potassium iodide</u>

<u>Calcium</u> + oxygen = Calcium oxide

Beryllium +<u> Bromine</u> = Beryllium bromide

<u>Copper + Oxygen</u> = Copper oxide

Explanation:

The balanced equation can be written as:

1.Potassium + Iodine =<u> Potassium iodide</u>

2K(s)+I_{2}(s)\rightarrow 2KI(s)

Here K = potassium

I2 = Iodine

KI = potasssium iodide.

2.<u>Calcium</u> + oxygen = Calcium oxide

2Ca(s)+O_{2}(g)\rightarrow 2CaO(s)

Ca = calcium

O2 = oxygen

CaO = Calcium Oxide

3.Beryllium +<u> Bromine</u> = Beryllium bromide

Be(s)+Br_{2}(g)\rightarrow BeBr_{2}(s)

Here,

Be = beryllium

Br2 = bromine

BeBr2 = Beryllium Bromide

4. Copper + Oxygen = Copper oxide

2Cu(s)+O_{2}(g)\rightarrow 2CuO(s)

Cu = Copper

O2 = oxygen

CuO = Copper Oxide

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If the solubility of KCl in 100 mL of H₂O is 34 g at 20 °C and 43 g at 50 °C, label each of the following solutions as unsaturat
sertanlavr [38]

Answer:

a) Unsaturated

b) Supersaturated

c) Unsaturated

Explanation:

A saturated  solution contains the <u>maximum amount of a solute that will dissolve in a given  solvent at a specific temperature</u>.

An unsaturated solution contains <u>less solute than it  has the capacity to dissolve. </u>

A supersaturated solution, <u>contains more  solute than is present in a saturated solution</u>. Supersaturated solutions are not very  stable. In time, some of the solute will come out of a supersaturated solution as crystals.

According to these definitions and considering that the solubility of KCl in 100 mL of H₂O at <u>20 °C is 34 g</u>, and at <u>50 °C is 43 g</u> we can label the solutions:

a) 30 g in 100 mL of H₂O at 20 °C  ⇒ unsaturated

b) 65 g in 100 mL of H₂O at 50 °C  ⇒ supersaturated

c) 42 g in 100 mL of H₂O at 50 °C and slowly cooling to 20 °C to give a clear solution <u>with no precipitate</u> ⇒ unsaturated (if it were saturated it would have had precipitate)

8 0
2 years ago
how many grams of phosphorus (P4) react with 35.5L of O2 at STP to form solid tetraphosphorus decaoxide
algol [13]

Answer:

mass P4 = 35.998 g

Explanation:

  • P4 + 5O2 → P4O10

∴ STP: P = 1 atm; T = 298 K

∴ V O2= 35.5 L

⇒ nO2 = P.V / R.T

∴ R = 0.082 atm.L/K.mol

⇒ nO2 = ((1 atm)×(35.5L))/((0.082 atm.L/K.mol)(298K))

⇒ nO2 = 1.453 mol O2

⇒ mol P4 = (1.453 molO2)×(mol P4/ 5molO2) = 0.2906 mol P4

∴ Mw P4 = 123.895 g/mol

⇒ mass P4 = (0.2906 mol P4)×(123.895 g/mol) = 35.998 g P4

4 0
3 years ago
A sample of a compound contains 60.0 g C and 5.05 g H. Its molar mass is 78.12 g/mol. What is the compound’s molecular formula
ki77a [65]
The standard atomic weight of a C is 12, and the standard atomic weight of a H is 1.  So to find molar ratio of C and H in the compound: 60.0/12=5, 5.05/1=5. This means the molar ratio of C and H is 5:5, thus 1:1.  Assuming the molecular formula is CnHn, to find molar mass: 12n + 1n = 78.12.  n=78.12/(12+1) = 6.  So the compound's molecular formula is C6H6, benzene.
4 0
2 years ago
Read 2 more answers
A 0.1375-g sample of solid magnesium is burned in a constant-volume bomb calorimeter that has a heat capacity of 3024 J/°C. The
Anna35 [415]

Answer:

-24.76 kJ/g; -601.8 kJ/mol

Explanation:

There are two heat flows in this experiment.

Heat from reaction + heat absorbed by calorimeter = 0

             q1                +                     q2                         = 0

           mΔH             +                    CΔT                        = 0

Data:

m = 0.1375 g

C = 3024 J/°C

ΔT = 1.126 °C

Calculations:

0.1375ΔH + 3024 × 1.126 = 0

           0.1375ΔH + 3405 = 0

                        0.1375ΔH = -3405

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ΔH = -24.76 kJ/g ×24.30 g/mol = -601.8 kJ/mol

3 0
3 years ago
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Contact [7]
Answer is: the nature of the initial nickel sulfide mixture is a suspension.
Suspension<span> is a </span>heterogeneous mixture (solute<span> particles do not </span>dissolve), <span>that contains </span>solid<span> particles (in this example nickel sulfide or NiS) sufficiently large for </span>sedimentation. <span> The internal phase (solid nickel sulfide) is dispersed throughout the external phase (water).</span>

8 0
3 years ago
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