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Rainbow [258]
4 years ago
9

What do the numbers listed for an element on the periodic table represent? The number with a lower value is the mass, and the nu

mber with a higher value is the atomic number. The number with a lower value is the number of protons, and the number with a higher value is the mass. The number with a lower value is the number of neutrons, and the number with a higher value is the number of electrons. The number with a lower value is the number of neutrons and protons, and the number with a higher value is the number of electrons.
Chemistry
1 answer:
attashe74 [19]4 years ago
7 0
The smaller number is the number of Protons *The Atomic Number*The greater number is the Atomic Mass. *Protons plus Neutrons*
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The concentration of sugar in a soft drink is measured to be 10.5%. how many grams of sugar are in 125 g of the drink?
viva [34]
The grams  of the sugar in 125 g of the drink is calculated as below

%M/m) = mass of the solute (sugar)/  mass of the  solvent(drink) x100

let the mass  of the solute(sugar) be represented by y

convert % into fraction by dividing by 100 = 10.5/100

10.5/100 = y/125

by cross multiplication

100y =1312.5
divide both side by 100

y=13.125  grams


5 0
4 years ago
For each image, select the state of matter represented.
kifflom [539]

Answer:

need image

Explanation:

5 0
3 years ago
Determine the oxidation state for each of the elements below. The oxidation state of ... silver ... in ... silver oxide Ag2O ...
KatRina [158]

Answer:

The oxidation state of silver in \rm Ag_2O is +1.

The oxidation state of sulfur in \rm SO_2 is +4.

Explanation:

The oxidation states of atoms in a compound should add up to zero.

<h3>Ag₂O</h3>

There are two silver \rm Ag atoms and one oxygen \rm O atom in one formula unit of \rm Ag_2O. Therefore:

\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}.

The oxidation state of oxygen in most compounds (with the exception of peroxides and fluorides) is -2. Silver oxide \rm Ag_2O isn't an exception. Therefore:

\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times (-2) = 0\end{aligned}.

Solve this equation for the (average) oxidation state of \rm Ag:

\text{Oxidation state of $\rm Ag$} = 1.

<h3>SO₂</h3>

Similarly, because there are one sulfur \rm S atom and two oxygen \rm O atoms in each \rm SO_2 molecules:

\begin{aligned}&\rm 1\times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}.

The oxidation state of \rm O in \rm SO_2 is also -2, not an exception, either.

Therefore:

\begin{aligned}&\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times (-2) = 0\end{aligned}.

Solve this equation for the oxidation state of \rm S here:

\text{Oxidation state of $\rm S$} = 4.

8 0
4 years ago
Metallurgy is the study of _____.
vovangra [49]
Hello,

The answer is "<span>behaviors and properties of metals".

Reason:

</span>The answer is the study of behaviors and properties of metals.

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit
5 0
4 years ago
Read 2 more answers
Hot lead with a mass of 200.0 g of (Specific heat of Pb = 0.129 J/gºC) at 176.4°C was dropped into a calorimeter containing an u
Morgarella [4.7K]

the Calorimetry relationships you can find the amount of water in the calorimeter is      m = 21.3 g

given parameters

  • Lead mass M = 200.0 g
  • Initial lead temperature T₁ = 176.4ºC
  • Specific heat of Lead    c_{e Pb} = 0.129 J / g ºC
  • Sspecific heat of water c_{e H_2O} = 4.186 J / g ºC
  • Initial water temperature T₀ = 21.7ºC
  • Equilibrium temperature T_f = 56.4ºC

to find

The body of water

Thermal energy is the energy stored in the body that can be transferred as heat when two or more bodies are in contact. Calorimetry is a technique where the energy is transferred between the body only in the form of heat and in this case the thermal energy of the lead is transferred to the calorimeter that reaches the equilibrium that the thematic energy of the two is equal

              Q_{ceded} = Q_{absorbed}

               

Lead, because it is hotter, gives up energy

              Q_{ceded} = M c_{e Pb} (T₁ - T_f)

The calorimeter that is colder absorbs the heat

              Q_{absrobed} = m c_{e H_2O} (T_f - T₀)

where M and m are the mass of lead and water, respectively, c are the specific heats, T₁ is the temperature of the hot lead, T₀ the temperature of cold water and T_f the equilibrium temperature

              M c_{ePb} (T₁ - T_f) = m c_{eH2O} (T_f - T₀)

               m = \frac{ M\  c_{ePb} \ (T_1 - T_f)}{c_{eH_2O}  \ (T_f - T_o)}

let's calculate

              m = \frac{200 \ 0.129  (176.4-56.4)}{ 4.186 \  (56.4 -21.7)}

              m = 3096 / 145.25

              m = 21.3 g

Using the Calorimetry relationships you can find the amount of water in the calorimeter is:

             m = 21.3 g

learn more about calorimetry here:

brainly.com/question/15073428

6 0
3 years ago
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