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s2008m [1.1K]
2 years ago
10

How many meters are in one yard? 1m=1.09​

Chemistry
2 answers:
GuDViN [60]2 years ago
4 0

Answer:

0.9144, hoped I helped

Darina [25.2K]2 years ago
3 0
0.91 meters are in one yard. If you wouldn’t mind I only need 2 more brainliests
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How do you determine the number of protons in an element
QveST [7]

Answer:

You determine the number of protons by its atomic number, just like electrons.

I am joyous to assist you anytime.

3 0
3 years ago
Solid potassium chlorate decomposes upon heating to form
olga55 [171]

Answer:

32.6%

Explanation:

Equation of reaction

2KClO₃ (s) → 2KCl (s) + 3O₂ (g)

Molar mass of 2KClO₃ = 245.2 g/mol ( 122.6 × 2)

Molar volume of Oxygen at s.t.p = 22.4L / mol

since the gas was collected over water,

total pressure = pressure of water vapor + pressure of  oxygen gas

0.976 = 0.04184211 atm + pressure of oxygen gas at 30°C

pressure of oxygen = 0.976 - 0.04184211 = 0.9341579 atm = P1

P2 = 1 atm, V1 = 789ml, V2 = unknown, T1 = 303K, T2 = 273k at s.t.p

Using ideal gas equation

\frac{P1V1}{T1} = \frac{P2V2}{T2}

V2 = \frac{P1V1T2}{T1P2}

V2 = 664.1052 ml

245.2   yielded 67.2 molar volume of oxygen

0.66411 will yield = \frac{245.2 * 0.66411}{67.2}  = 2.4232 g

percentage of potassium chlorate in the original mixture = \frac{2.4232 * 100}{7.44} = 32.6%

3 0
3 years ago
Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e
notsponge [240]

<u>Answer:</u> The concentration of Sn^{2+} in the cell is 9.0\times 10^{-3}M

<u>Explanation:</u>

We are given:

<u>Oxidation half reaction:</u>  Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-   E^o_{Zn^{2+}/Zn}=-0.76V

<u>Reduction half reaction:</u>  Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)   E^o_{Sn^{2+}/Sn}=-0.136V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

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

Putting values in above equation, we get:

E^o_{cell}=-0.136-(-0.76)=0.624V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}

where,

E_{cell} = electrode potential of the cell = 0.660 V

E^o_{cell} = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

[Zn^{2+}]=2.5\times 10^{-3}M

[Sn^{2+}] = ?

Putting values in above equation, we get:

0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})

[Sn^{2+}]=9.0\times 10^{-3}M

Hence, the concentration of Sn^{2+} ions is 9.0\times 10^{-3}M

3 0
3 years ago
What action would you expect an atom of Aluminum to undergo?
KATRIN_1 [288]

Answer:

C. Lose three electrons to have a full outer shell  

Explanation:

Al is in Group 13 of the Periodic Table, so it has three valence electrons.

It must either lose three electrons or gain five to achieve a stable octet.

It is easier to lose three electrons than it is to gain five, so Al loses three electrons.

D. is wrong, for the same reason.

A. is wrong. If Al lost three electrons, it would be breaking into a stable inner shell.

C. is wrong. Al is a metal, so it will lose electrons in a reaction.

6 0
3 years ago
What is the pOH of<br> a<br> 2.6 x 10-6 M H+ solution?
scZoUnD [109]

Answer:

pH = -log 2.6 x 10-6 M

pH = 5.585

pOH= 14 - 5.585 = 8.415

8.4 as 2 sig figs

8 0
3 years ago
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