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3 years ago

Which of the following statements about atoms is NOT true?

Chemistry

1 answer:

Alika [10]3 years ago

6 0

Answer:

atoms are too small to be seen under a microscope

Explanation:

scientists have been always making research on atomic structure which cannot be directly looked through a microscope

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How many moles are in 175.9 grams of Pb3(PO4)2?

snow_lady [41]

Answer:

0.21674767727138733

Explanation:

Again I don't want to give an explanation right now because I can't exactly do that more than once. it hurts my child brain. And the fact that I just woke up.

8 0

3 years ago

A 100. mL volume of 0.800 M calcium chloride is mixed with 400. mL of water to make 500. mL of solution. What is the final molar

strojnjashka [21]

Answer:

0.99mol/L

Explanation:

Molarity ( M) = # of moles of solute / volume of solution (L)

Volume of Solution = 100 mL or 0.1 L

Molarity ( M) = # of moles of solute / volume of solution (L)

Molarity (M) = 0.099 mol / 0.1 L = 0.99mol/ L

3 0

3 years ago

In thermodynamics, we determine the spontaneity of a reaction by the sign of ΔG. In electrochemistry, spontaneity is determined

faltersainse [42]

Answer:

For A: The standard cell potential of the reaction is 4.4 V

For B: The standard Gibbs free energy of the reaction is $-8.50\times 10^5J$

For C: The reaction is spontaneous as written.

Explanation:

For A:

The given chemical reaction follows:

$2Li(s)+Cl_2(g)\rightarrow 2Li^+(aq.)+2Cl^-(aq.)$

The given half reaction follows:

Oxidation half reaction: $Li(s)\rightarrow Li^+(aq.)+e^-;E^o_{Li^+/Li}=-3.04V$ ( × 2)

Reduction half reaction: $Cl_2(g)+2e^-\rightarrow 2Cl^-(aq.);E^o_{Cl_2/2Cl^-}=+1.36V$

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

Here, chlorine will undergo reduction reaction will get reduced.

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:

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

$E^o_{cell}=1.36-(-3.04)=4.4V$

Hence, the standard cell potential of the reaction is 4.4 V

For B:

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

where,

n = number of electrons transferred = $2mol\text{ e}^-$

F = Faradays constant = $96500J/V.mol\text{ e}^-$

$E^o_{cell}$ = standard cell potential = 4.4 V

Putting values in above equation, we get:

$\Delta G^o=-2\times 96500\times 4.4=-849200J=-8.50\times 10^5J$

Hence, the standard Gibbs free energy of the reaction is $-8.50\times 10^5J$

For C:

For a reaction to be spontaneous, the standard Gibbs free energy change of the reaction must be negative.

From above, the standard Gibbs free energy change of the reaction is coming out to be negative.

Hence, the reaction is spontaneous as written.

7 0

3 years ago

Potassium iodide reacts with lead(II) nitrate in this precipitation reaction: 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s) Wha

andrew11 [14]

Answer:

a. 174 mL

Explanation:

Let's consider the following reaction.

2 KI(aq) + Pb(NO₃)₂(aq) → 2 KNO₃(aq) + PbI₂(s)

We have 155.0 mL of a 0.112 M lead(II) nitrate solution. The moles of Pb(NO₃)₂ are:

0.1550 L × 0.112 mol/L = 0.0174 mol

The molar ratio of KI to Pb(NO₃)₂ is 2:1. The moles of KI are:

2 × 0.0174 mol = 0.0348 mol

The volume of a 0.200 M KI solution that contains 0.0348 moles is:

0.0348 mol × (1 L / 0.200 mol) = 0.174 L = 174 mL

5 0

3 years ago

HELP ILL GIVE U FREE POINTS AND BRAINLEST !! Observe the plate boundary shown here. What type of plate boundary is this? Many ge

Studentka2010 [4]

The answer is B.
You can rule C out because divergent means moving away. Rule out A because there is an oceanic and continental plate, not 2 of the same type. Rule D out for the same reason.

3 0

3 years ago

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