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2 years ago

A chemist is studying how a new type of paper absorbs ink. Which is a macroscopic view of the chemist’s subject?

Chemistry

2 answers:

nadezda [96]2 years ago

8 0

B. one sheet of the new type of paper

dalvyx [7]2 years ago

8 0

Answer: B. one sheet of the new type of paper.

Explanation: Macroscopic views are the those observable which are presented on the larger scale rather than being on the smaller scale.

Now, it is given that the chemist us studying a new type pf paper that absorbs ink.

Thus option A explains the microscopic view about the individual fibers that make up the paper. Those small fibers which are responsible for the making of the paper.

Option C also explains the microscopic view as it is asking for the chemical composition of the fibers of the paper.

Thus option B is the only macroscopic view which explains the whole one sheet of the new type of water which includes the knowledge of the individual fibers as well as the chemical composition of the fibers as well.

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Select all that apply.

dem82 [27]

For the answer to the question above, those that apply are: 
- salt is not chemically bonded to water 
- the ratio of salt to water may vary 
- salt and water retain their own chemical properties
I hope my answer helped you

6 0

2 years ago

Use bond energies to calculate ΔHrxn Δ H r x n for the reaction. 2H2(g)+O2(g)→2H2O(g) 2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( g )

Olenka [21]

Answer:

$\large \boxed{\text{-486 kJ}}$

Explanation:

You calculate the energy required to break all the bonds in the reactants.

Then you subtract the energy needed to break all the bonds in the products.

2H₂ + O₂ ⟶ 2H-O-H

Bonds: 2H-H 1O=O 4H-O

D/kJ·mol⁻¹: 436 498 464

$\begin{array}{rcl}\Delta H & = & \sum{mD_{\text{reactants}}} - \sum{nD_{\text{products}}}\\& = & 2 \times 436 +1 \times 498 - 4 \times 464\\&=& 1370 - 1856\\&=&\textbf{-486 kJ}\\\end{array}\\\text{The enthalpy of reaction is $\large \boxed{\textbf{-486 kJ}}$}.$

3 0

2 years ago

Where photosynthesis takes place?

Vlad [161]

Answer:

chloroplasts

Explanation:

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]

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

Explanation:

We are given:

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

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

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

If 10.00 g of iron metal is burned in the presence of excess of O2 how many grams of Fe2O3 will form

sergiy2304 [10]

14.292 grams of Fe2O3 is formed when 10 gram of iron metal is burned.

Explanation:

The balanced equation for the reaction is to be known so that number of moles taking part can be known.

The balanced chemical equation is

4Fe + 3 $O_{2}$ ⇒ 2 $Fe{2}$ $O{3}$

From the given weight of iron to be used for the production of $Fe{2}$ $O{3}$ , number of moles of Fe taking part in the reaction can be known by the formula:

Number of moles= mass ÷ Atomic mass of one mole of the element.

(Atomic weight of Fe is 55.845 gm/mole)

Putting the values in equation

Number of moles = 10 gm ÷ 55.845 gm/mole

= 0.179 moles

Applying the stoichiometry concept

4 moles of Fe gives 2 Moles of Fe2O3

0.179 moles will produce x moles of Fe2O3

So, 2÷ 4 = x ÷ 0.179

2/4 = x/ 0.179

2 × 0.179 = 4x

2 × 0.179 / 4 = x

x = 0.0895 moles

So from 10 grams of iron metal 0.0895 moles of Fe2O3 is formed.

Now the formula used above will give the weight of Fe2O3

weight = atomic weight × number of moles

= 159.69 grams × 0.0895

= 14.292 grams of Fe2O3 formed.

4 0

3 years ago