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iren2701 [21]
2 years ago
9

A scientist observes that the electrical resistance of a superconducting material drops to zero when the material is cooled

Chemistry
1 answer:
frutty [35]2 years ago
7 0

Answer:

<h2>The scientist is observing an intensive property of a superconductor.</h2>

Explanation:

This is the answer because no matter how much of the superconductor material is present  the resistance of the superconducting material still drops to zero when the material is cooled to very low temperatures

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Why do we slant the graduated cylinder before dropping the metal cylinder into it?
alexandr402 [8]

So the solution inside doesn't splash

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Two students made the Lewis dot diagrams of H2O. The diagrams are as shown.
netineya [11]

Answer:

Only student A

Explanation:

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If parents wanted to pick their child's traits like hair color, gender, eye color, what would doctors have to do?
kap26 [50]

Answer:

Nothing

Explanation:

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5 0
3 years ago
How to calculate delta S surroundings? calculate Delta S(surr) at the indicated temperature for a reaction having each of the fo
astraxan [27]
You can use the equation ΔS(surr)=q(surr)/T or ΔS(surr)=-q(rxn)/T.
the two equations are equal since we know that the energy the system (reactoin) puts out just goes into the surroundings.  
(In other words q(surr)=-q(rxn))

Using the equation, <span>ΔS(surr)=-(-283kJ/298K)=0.9497kJ/K or 949.7J/K

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8 0
3 years ago
In a single displacement reaction between sodium phosphate and barium, how much of each product (in grams) will be formed from 1
weeeeeb [17]

Answer:

A. 3.36g of Na.

B. 14.62g of Ba3(PO4)2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

3Ba + 2Na3PO4 → 6Na + Ba3(PO4)2

Next, we shall determine the mass of Ba that reacted and the mass of Na and Ba3(PO4)2 produced from the equation.

This is illustrated below:

Molar Mass of Ba = 137g/mol

Mass of Ba from the balanced equation = 3 x 137 = 411g

Molar mass of Na = 23g/mol

Mass of Na from the balanced equation = 6 x 23 = 138g

Molar mass of Ba3(PO4)2 = (3 x 137) + 2[31 + (4x16)] = 411 + 2[31 + 64] = 601g/mol

Mass of Ba3(PO4)2 from the balanced equation = 1 x 601 = 601g

Summary:

From the balanced equation above,

411g of Ba reacted to produce 138g of Na and 601g of Ba3(PO4)2.

A. Determination of the mass of Na produced by reacting 10g of Ba.

From the balanced equation above,

411g of Ba reacted to produce 138g of Na.

Therefore, 10g of Ba will react to produce = (10 x 138)/411 = 3.36g of Na.

Therefore, 3.36g of Na is produced.

B. Determination of the mass of Ba3(PO4)2 produced by reacting 10g of Ba.

From the balanced equation above,

411g of Ba reacted to produce 601g of Ba3(PO4)2.

Therefore, 10g of Ba will react to produce = (10 x 601)/411 = 14.62g of Ba3(PO4)2.

Therefore, 14.62g of Ba3(PO4)2 is produced.

7 0
3 years ago
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