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Dafna1 [17]
2 years ago
7

There are the same number of these two particles in an atom

Chemistry
1 answer:
seropon [69]2 years ago
6 0

Answer:

Protons and electrons

Explanation:

Atoms that have a neutral charge have an equal number of protons and electrons. Protons are positively charged and electrons are negatively charged, so to cancel out all charges, the number of both these particles must be equal. If there is an unequal amount, then it is called an ion. Ions are atoms that have either a negative or positive charge because the amount of protons and electrons is not balanced.

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The given potentials are observed for a calomel electrode: E ° = 0.268 V and E ( saturated KCl ) = 0.241 V . Use this informatio
Anvisha [2.4K]

Answer: The E for Silver-silver Chloride electrode = 0.287 V

Explanation:

Silver/Silver Chloride (Ag/AgCl) with a value for E° that is actually +0.222 V or approximately 0.23 V has the actual potential of the half-cell prepared in this way as +0.197 V vs SHE, (Standard Hydrogen Electrode) which arises because in addition to KCl, there is the contribuion of AgCl to the chloride activity, which isn't exactly unity.

Therefore, the E for the Ag/AgCl electrode would approximately equal 0.287 V

4 0
3 years ago
Which of the following best describes careers that use chemistry?
telo118 [61]

Answer:

D

Explanation:

7 0
3 years ago
Using the data below, calculate the enthalpy for the combustion of C to CO C(s) + O2(g)--> CO2 (g) ΔH1 = -393.5 kJ CO(g) + 1/
natali 33 [55]

Answer:

ΔH = -110.5kJ

Explanation:

It is possible to obtain enthalpy of combustion of a particular reaction by the algebraic sum of similar reactions (Hess's law). Using:

1. C(s) + O₂(g) → CO₂(g) ΔH₁ = -393.5kJ

2. CO(g) + 1/2O₂(g) → CO₂(g) ΔH₂ = -283.0kJ

The sum of 1 -2 gives:

C(s) + <u>O₂(g)</u> + <u>CO₂(g)</u> → <u>CO₂(g)</u> + CO(g) + <u>1/2O₂(g)</u>

C(s) + 1/2O₂(g) → CO(g) ΔH = -393.5kJ - (-283.0kJ) =

<h3>ΔH = -110.5kJ</h3>

<u />

5 0
3 years ago
Give an example of organic and inorganic compound
Olenka [21]
Organic: sugar
inorganic: salt
5 0
4 years ago
The practical limit to ages that can be determined by radiocarbon dating is about 41000-yr-old sample, what percentage of the or
Aloiza [94]

Answer:

In percentage, the sample of C-4 remains = 0.7015 %

Explanation:

The Half life  Carbon 14 =  5730 year

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

k=\frac {ln\ 2}{5730}\ hour^{-1}

The rate constant, k = 0.000120968 year⁻¹

Time = 41000 years

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

So,  

\frac {[A_t]}{[A_0]}=e^{-0.000120968\times 41000}

\frac {[A_t]}{[A_0]}=0.007015

<u>In percentage, the sample of C-4 remains = 0.7015 %</u>

3 0
3 years ago
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