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Talja [164]
3 years ago
11

Helium decays to form lithium. Which equation correctly describes this decay? Superscript 6 Subscript 2 Baseline Upper H e right

arrow Superscript 3 Subscript 3 Baseline Upper L i + Superscript 0 Subscript negative 1 Baseline e Superscript 6 Subscript 2 Baseline Upper H e right arrow Superscript 6 Subscript 3 Baseline Upper H + Superscript 0 Subscript negative 1 Baseline e Superscript 6 Subscript 2 Baseline Upper H e right arrow Superscript 6 Subscript 3 Baseline Upper L i + Superscript 0 Subscript negative 1 Baseline e Superscript 6 Subscript 2 Baseline Upper H e right arrow Superscript 6 Subscript 3 Baseline Upper L i + Superscript 0 Subscript 1 Baseline e
Chemistry
2 answers:
Virty [35]3 years ago
4 0

Answer:

Superscript 6 Subscript 2 Baseline Upper H e right arrow Superscript 6 Subscript 3 Baseline Upper L i + Superscript 0 Subscript negative 1 Baseline e

Explanation:

Helium decays by beta emission, in beta emission, the mass number of the parent and daughter nuclei are the same but the atomic number of the daughter nucleus increases by one hence it is found one place to the right of its parent in the periodic table.

Having said this, the decay of Helium to lithium may be written as follows;

6/2He -> 6/3Li + 0/-1e

nirvana33 [79]3 years ago
4 0

Answer:

C

Explanation:

Edge 2020

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A carpet sells for $28.99 a square yard. What is the price of the carpet per square meter?
9966 [12]

It is given that the price of the carpet is $ 28.99 per square yard

Step 1: Convert square yard to square meter

1 square yard = 0.836 square meter

Step 2: Calculate the cost/square meter

Now based on the relation in step (1) we can write

0.836 square meter of carpet costs $ 28.99

Therefore the cost of 1 square meter of the carpet would be

= $ 28.99 * 1 sq meter/0.836 sq meter

= $ 34.68

7 0
4 years ago
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A 2.684-g sample of zinc oxide was reduced by hydrogen gas, resulting in 2. 156 g of pure zinc metal. Determine the empirical fo
Afina-wow [57]

The empirical formula of the initial zinc oxide is ZnO.

<h3>What is Empirical Formula?</h3>

The empirical formula of a compound represents the ratios of elements in a compound but not the actual numbers or arrangement of the atoms.

It is the lowest whole number ratio of the element in the compound.

<h3>How to find out the empirical formula?</h3>
  • Find out the given masses and molar masses of the elements

The molar mass of Zn = 65 gmol⁻¹

Given the mass of Zn = 2.156 g

The molar mass of Oxygen = 16 gmol⁻¹

The mass of Oxygen = Mass of a sample of zinc oxide - the mass of zinc metal

                                   = (2.684 - 2.156) g

                                   = 0.528 g

  • Find the number of moles of the elements in the compound

The number of moles is given by

n = \frac{m}{M}

where m = given mass and

M = Molar mass

Number of moles of Zinc = \frac{2.156}{65} = 0.033 moles

Number of moles of Oxygen =\frac{0.528}{16} = 0.033 moles

  • Find the simplest ratios of the elements in the compound. To find the ratios simply divide the number of moles by the lowest number of moles obtained.

Here, the number of moles is the same for both elements. Hence, the simplest ratio for Zn:O is 1:1.

Therefore, the empirical formula of zinc oxide is ZnO.

Learn more about the empirical formula:

brainly.com/question/1603500

#SPJ4

 

5 0
1 year ago
The pka of hf is 3.2 determine the pkb of hf?
Julli [10]

Well, first we must remember that

pK_{a}+pK_{b}=14

This is because

K_{a}*K_{b}=10^{-14}

-log(K_{a}*K_{b})=-log(10^{-14})\\-logK_{a}+-logK_{b}=-log(10^{-14})\\pK_{a}+pK_{b}=14

So then

pK_{b}=14-pK_{a}=14-3.2=1.8

7 0
4 years ago
Read 2 more answers
Calculate for the following electrochemical cell at 25°C, Pt H2(g) (1.0 atm) H (0.010 M || Ag (0.020 M) Ag if E (H) - +0.000 V a
viva [34]

Answer : The correct option is, (b) +0.799 V

Solution :

The values of standard reduction electrode potential of the cell are:

E^0_{[H^{+}/H_2]}=+0.00V

E^0_{[Ag^{+}/Ag]}=+0.799V

From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : H_2\rightarrow 2H^{+}+2e^-    

Reaction at cathode (reduction) : Ag^{+}+1e^-\rightarrow Ag    

The balanced cell reaction will be,  

H_2+2Ag^{+}\rightarrow 2H^{+}+2Ag

Now we have to calculate the standard electrode potential of the cell.

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}

E^o_{cell}=(+0.799V)-(+0.00V)=+0.799V

Therefore, the standard cell potential will be +0.799 V

4 0
3 years ago
If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.
vesna_86 [32]

Answer:

Sr(OH)2

Explanation:

We'll begin by calculating the number of mole of carbonic acid in 150mL of 3.5 M carbonic acid solution. This is illustrated below:

Molarity = 3.5M

Volume = 150mL = 150/1000 = 0.15L

Mole of carbonic acid, H2CO3 =..?

Mole = Molarity x Volume

Mole of carbonic acid, H2CO3 = 3.5 x 0.15 = 0.525 mole.

Next, we shall convert 0.525 mole of carbonic acid, H2CO3 to grams.

Mole of H2CO3 = 0.525 mole

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 =..?

Mass = mole x molar mass

Mass of H2CO3 = 0.525 x 62 = 32.55g

Next, we shall write the balanced equation for the reaction. This is given below:

Sr(OH)2 + H2CO3 → SrCO3 + 2H2O

Next, we shall determine the mass of Sr(OH)2 and H2CO3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Sr(OH)2 = 88 + 2(16 + 1) = 88 + 2(17) = 122g/mol

Mass of Sr(OH)2 from the balanced equation = 1 x 122 = 122g

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 from the balanced equation = 1 x 62 = 62g.

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Therefore, 12.5g of Sr(OH)2 will react with = (12.5 x 62)/122 = 6.35g.

We can see evidently from the calculations made above that it will take 6.35g out 32.55g of H2CO3 to react with 12.5g of Sr(OH)2. Therefore, Sr(OH)2 is the limiting reactant and H2CO3 is the excess reactant

5 0
3 years ago
Read 2 more answers
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