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Sever21 [200]
2 years ago
14

In the compound MgCl2MgCl2, the subscript 2 indicates that

Chemistry
1 answer:
MatroZZZ [7]2 years ago
3 0

 The subscript 2  indicate that  there are two atoms of chlorine  in the compound.

<u><em> Explanation</em></u>

 Subscript in the chemical formula indicate  the number of atoms  of the element immediately  before the subscript.

In MgCl₂  there are two(2)  atoms of chlorine  since Subscript 2 is immediately  after chlorine.

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What is the energy of a photon of green light whose frequency is 6.85 X 1014/seC?
Anvisha [2.4K]

Frequency of photon = is 6.85 X 10¹⁴ sec⁻¹

Energy of photon , E = hv

where h is Planck's constant, v is the frequency of photon

E = 6.63 × 10⁻³⁴ J.s x 6.85 X 10¹⁴ s⁻¹

E = 4.54 x 10⁻¹⁹ J

Therefore, the energy of a photon of green light is 4.54 x 10⁻¹⁹ J.

8 0
2 years ago
the molar enthalpy of formation of carbondioxide is -393kjmol. calculate the heat released by the burning of 0.327g of carbon to
Arte-miy333 [17]

This is equivalent to having a standard enthalpy change of reaction equal to  10.611 kJ

<u>Explanation</u>:

The standard enthalpy change of reaction,  Δ H ∘ , is given to you in kilojoules per mole, which means that it corresponds to the formation of one mole of carbon dioxide.

                                    C (s]  +  O 2(g] → CO 2(g]

Remember, a negative enthalpy change of reaction tells you that heat is being given off, i.e. the reaction is exothermic.

First to convert  grams of carbon into moles,

use carbon's molar mass(12.011 g).

                    Moles of C = mass in gram / molar mass

                                        = 0.327 g  / 12.011 g

                     Moles of C = 0.027 moles

Now, in order to determine how much heat is released by burning of 0.027 moles of carbon to form carbon-dioxide.

                                        =  0.027 moles C  \times 393 kJ

             Heat released  = 10.611  kJ.

So, when  0.027  moles of carbon react with enough oxygen gas, the reaction will give off  10.611 kJ  of heat.

This is equivalent to having a standard enthalpy change of reaction equal to  10.611 kJ

7 0
3 years ago
Potassium (K) and calcium fluoride (CaF2) can both be classified as<br> (Please help)
nasty-shy [4]
Potassium and calcium fluoride are both metals
8 0
3 years ago
You’re going on a plane from Alaska to Florida and you’re taking your bike. It’s the middle of the winter. You fill up your tire
12345 [234]

Answer:

5.52atm

Explanation:

Using the pressure law formula:

P1/T1 = P2/T2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the question, the following information were provided;

P1 = 4.72 atm

P2 = ?

T1 = -3.50°C = -3.50 + 273 = 269.5K

T2 = 42°C = 42 + 273 = 315K

Using P1/T1 = P2/T2

4.72/269.5 = P2/315

CROSS MULTIPLY

4.72 × 315 = 269.5 × P2

1,486.8 = 269.5P2

P2 = 1,486.8 ÷ 269.5

P2 = 5.52atm

8 0
2 years ago
A concentration cell is constructed by using the same half-reaction for both the cathode and anode. What is the value of standar
Akimi4 [234]

Solution :

A cell that is concentrated is constructed by the same half reaction for the anode as well as he cathode.

We know,

In a standard cell,

the reduction half cell reaction is :

$Ag^+(aq)+e^- \rightarrow Ag(s) E^0 = -0.80 \ V$

The oxidation half ell reaction :

$Ag(s) \rightarrow Ag^+(aq) + e^- \ E^0= +0.80 \ V$

Thus the complete reaction of the cell is :

$Ag^+(aq)+ Ag(s) \rightarrow Ag^+(aq)+Ag(s)$

$E^0 $ cell = $E_R - E_L = 0.00  \ \text{volts}$

7 0
2 years ago
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