4. Will the net force be balanced or unbalanced? 7N 4N A. Balanced B. Unbalanced

Why did scientists in the 1800s decide that the list of known elements needed to be more

ankoles [38]

Explanation:

The scientists realized that the properties of elements related to each other depending on relational properties of their atoms. They realized that atoms of different elements shared some properties depending on the number of electrons they have in their valence orbital shells and atomic mass numbers. These atomic properties determined the elements' shared properties such as texture, reactivity, electric conductivity and etcetera. Arranging the atoms in a periodic table would therefore ideally make it easier to decipher the shared similarities and differences between atoms of different elements, such as physical and chemical properties.

Learn More:

For more on the periodic table check out these questions;

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8 0

3 years ago

Who discovered the flame test

Genrish500 [490]

Answer:

Thomas Melvill

Explanation:

...

6 0

3 years ago

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,

Masja [62]

Answer: The concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

Explanation:

The given cell is:

$Pt(s)|H_2(g.1atm)|H^+(aq.,1.0M)||Au^{3+}(aq,?M)|Au(s)$

Half reactions for the given cell follows:

Oxidation half reaction: $H_2(g)\rightarrow 2H^{+}(1.0M)+2e^-;E^o_{H^+/H_2}=0V$ ( × 3)

Reduction half reaction: $Au^{3+}(?M)+3e^-\rightarrow Au(s);E^o_{Au^{3+}/Au}=1.50V$ ( × 2)

Net reaction: $3H_2(s)+2Au^{3+}(?M)\rightarrow 6H^{+}(1.0M)+2Au(s)$

Substance getting oxidized always act as anode and the one getting reduced always act as 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:

$E^o_{cell}=1.50-0=1.50V$

To calculate the concentration of ion for given EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^6}{[Au^{3+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = 1.23 V

$E^o_{cell}$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[Au^{3+}]=?M$

$[H^{+}]=1.0M$

Putting values in above equation, we get:

$1.23=1.50-\frac{0.059}{6}\times \log(\frac{(1.0)^6}{[Au^{3+}]^2})$

$[Au^{3+}]=1.87\times 10^{-14}M$

Hence, the concentration of $Au^{3+}$ in the solution is $1.87\times 10^{-14}M$

7 0

3 years ago

What is the percent of nitrogen in dinitrogen pentoxide?

Lostsunrise [7]

Answer: D.) 25.9%

Explanation:

Dinitrogen pentoxide chemical formular : N2O5

Calculating the molar mass of N2O5

Atomic mass of nitrogen(N) = 14

Atomic mass of oxygen(O) = 16

Therefore molar mass :

N2O5 = (2 × 14) + (5 × 16) = 28 + 80 = 108g/mol

Percentage amount of elements in N205:

NITROGEN (N) :

(Mass of nitrogen / molar mass of N2O5) × 100%

MASS OF NITROGEN = (N2) = 2 × 14 = 28

PERCENT OF NITROGEN : (28/108) × 100%

0.259259 × 100%

= 25.925%

= 25.9%

4 0

3 years ago

What is the formula for the compound formed by calcium ions and chloride ions? (1) CaCl (2) CaCl2 (3) CaCl3 (4) Ca2Cl

amid [387]

I think it would be c

6 0

3 years ago