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Karolina [17]
3 years ago
14

How is my crime scene sketches? What grade are these going to get?

Chemistry
2 answers:
NeTakaya3 years ago
4 0

Answer:

Yeah very good job maybe a 100%.

andre [41]3 years ago
3 0

Good job! Could get a 100

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Phosphorous pentoxide, P2O5(s), is produced from the reaction between pure oxygen and pure phosphorous (P, solid). What is the v
BartSMP [9]

Answer:

4190.22 L = 4.19 m³.

Explanation:

  • For the balanced reaction:

<em>2P₂ + 5O₂ ⇄ 2P₂O₅. </em>

It is clear that 2 mol of P₂ react with <em>5 mol of O₂ </em>to produce <em>2 mol of P₂O₅.</em>

  • Firstly, we need to calculate the no. of moles of 6.92 kilograms of P₂O₅ produced through the reaction:

no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.

  • Now, we can find the no. of moles of O₂ is needed to produce the proposed amount of P₂O₅:

<u><em>Using cross multiplication:</em></u>

5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.

??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.

∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.

  • Finally, we can get the volume of oxygen using the general law of ideal gas:<em> PV = nRT.</em>

where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).

V is the volume of the gas in L (V = ??? L).

n is the no. of moles of the gas in mol (n = 60.95 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).

∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.

3 0
3 years ago
For the reaction shown, identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent. KNO3 →
Vinvika [58]

Answer : The oxidizing element is N and reducing element is O. 

KNO_{3} is act as an oxidizing agent as well as reducing agent.

Explanation :

An Oxidizing agent is the agent which has ability to oxidize other or a higher in oxidation number.

Reducing agent is the agent which has ability to reduce other or lower in oxidation number.

The given reaction is :

KNO_{3} \rightarrow KNO_{2} +O_{2}

KNO_{3}  act as an oxidizing agent.

The oxidation number of N in KNO_{3} is calculated as:

(+1)+(x)+3(-2) = 0

x = +5

And the oxidation number of N in KNO_{2}  is calculated as:

(+1)+(x)+2(-2) = 0

x = +3

From the oxidation number method, we conclude that the oxidation number  reduced this means KNO_{3} itself get reduced to KNO_{2} and it can act as an oxidizing agent.

KNO_{3}  act as a reducing agent.

KNO_{3} \rightarrow KNO_{2} +O_{2}

The oxidation number of O in KNO_{3} is calculated as:

(+1)+(+5)+3(x) = 0

x = -2

The oxidation number of O in O_{2} is Zero (o).

Now, we conclude that the oxidation number increases this means KNO_{3} itself get oxidized to O_{2} and it can act as reducing agent.





                     

4 0
3 years ago
Read 2 more answers
Why polystyrene containers are used to store ice cubes?​
ASHA 777 [7]

Answer:

Hope this may help you

6 0
2 years ago
Three players (A, B and C) shoot three arrows at a target
UkoKoshka [18]

Answer:

Player B

Explanation:

I just did it and I got it right :)

6 0
3 years ago
Hard water often contains dissolved Ca2+ and Mg2+ ions. One way to soften water is to add phosphates. The phosphate ion forms in
avanturin [10]
<span>5.5×10−2M in calcium chloride and 8.0×10−2M in magnesium nitrate.
What mass of sodium phosphate must be added to 1.5L of this solution to completely eliminate the hard water ion

1) Content of Ca (2+) ions

Calcium chloride = CaCl2

Ionization equation: CaCl2 ---> Ca (2+) + 2 Cl (-)

=> Molar ratios: 1 mol of CaCl2 : 1 mol Ca(2+) : 2 mol Cl(-)

Calculate the number of moles of CaCl2 in 1.5 liters of 5.5 * 10^-2 M solution

M = n / V => n = M*V = 5.5 * 10^ -2 M * 1.5 l = 0.0825 mol CaCl2

=> 0.0825 mol Ca(2+)

2) Number of phosphate ions needed to react with 0.0825 mol Ca(2+)

formula of phospahte ion: PO4 (3-)

molar ratio: 2PO4(3-) + 3Ca(2+) = Ca3 (PO4)2

Proportion: 2 mol PO4(3-) / 3 mol Ca(2+) = x / 0.0825 mol Ca(2+)

=> x = 0.0825 coml Ca(2+) * 2 mol PO4(3-) / 3 mol Ca(2+) = 0.055 mol PO4(3-)

3) Content of Mg(2+) ions

Ionization equation: Mg (NO3)2 ----> Mg(2+) + 2 NO3 (-)

Molar ratios: 1 mol Mg(NO3)2 : 1 mol Mg(2+) + 2 mol NO3(-)

number of moles of Mg(NO3)2 in 1.5 liter of 8.0 * 10^-2 M solution

n = M * V = 8.0 * 10^ -2 M * 1.5 liter = 0.12 moles Mg(NO3)2

ions of Mg(2+) = 0.12 mol Mg(NO3)2 * 1 mol Mg(2+) / mol Mg(NO3)2 = 0.12 mol Mg(2+)

4) Number of phosphate ions needed to react with 0.12 mol Mg(2+)

2PO4(3-) + 3Mg(2+) = Mg3(PO4)2

=> 2 mol PO4(3-) / 3 mol Mg(2+) = x / 0.12 mol Mg(2+)

=> x = 0.12 * 2/3 mol PO4(3-) = 0.16 mol PO4(3-)

5) Total number of moles of PO4(3-)

0.055 mol + 0.16 mol = 0.215 mol

6) Sodium phosphate

Sodium phosphate = Na3(PO4)

Na3PO4 ---> 3Na(+) + PO4(3-)

=> 1 mol Na3PO4 : 1 mol PO4(3-)

=> 0.215 mol PO4(3-) : 0.215 mol Na3PO4

mass in grams = number of moles * molar mass

molar mass of Na3 PO4 = 3*23 g/mol + 31 g/mol + 4*16 g/mol = 164 g/mol

=> mass in grams = 0.215 mol * 164 g/mol = 35.26 g

Answer: 35.26 g of sodium phosphate
</span>
5 0
3 years ago
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