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3 years ago

Which state of matter has particles that vibrate in a rigid structure

Chemistry

1 answer:

dedylja [7]3 years ago

4 0

Explanation:

Solids have definite shape and distinct boundaries as such the intermolecular forces between them is maximum

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the atomic number of an element x is 14 and its relative atomic mass is 28 write the electronic distribution in an atom of x

AlekseyPX

Answer:

the electronic distribution in an atom of x is 14

3 0

2 years ago

Would a false positive from the reaction between the inoculating loop and hydrogen peroxide

Svetradugi [14.3K]

The false positive from the response of hydrogen peroxide and the immunizing circle would be created by poor specificity. The recipe for specificity is TN/TN+FP. False-positive outcomes can be ascribed to meddling substances in nature where the strips are put away or utilized, for example, hydrogen peroxide (H2O2) or fade (hypochlorite).

3 0

3 years ago

How long would it take to produce enough aluminum to make a case (24 cans) of aluminum soft drink cans if each can used 3 g of a

kozerog [31]

Answer : It takes time to produce 3 g of aluminium is, 165 seconds.

Explanation :

As we are given that the mass of aluminium is, 3 grams. Now we have to calculate the total mass of aluminum.

Total mass of aluminum = 3g × 24 = 72 g

Now we have to calculate the moles of aluminum.

$\text{Moles of Al}=\frac{\text{Mass of Al}}{\text{Molar mass of Al}}$

Molar mass of Al = 27 g/mol

$\text{Moles of Al}=\frac{72g}{27g/mol}=2.67mol$

As, 1 mole of Al has 3 Faradays

So, 2.67 mole of Al has = $2.67\times 3=80.1$ Faradays

and,

Charge = $80.1\text{ Faradays}\times \frac{96500\text{ coulombs}}{1\text{ Faradays}}=7.73\times 10^6\text{ coulombs}$

Current = $5.00\times 10^4A\times \frac{93.9}{100}=46950A$

Now we have to calculate the time.

$Time=\frac{Charge}{Current}=\frac{7.73\times 10^6}{46950}=164.6s\aaprox 165s$

Hence, it takes time to produce 3 g of aluminium is, 165 seconds.

5 0

3 years ago

Help please<br>........

IgorLugansk [536]

I am pretty sure it is Vitamin A... but I could be wrong. Sorry if it is wrong.

3 0

2 years ago

A 57.07 g sample of a substance is initially at 24.3°C. After absorbing of 2911 J of heat, the temperature of the substance is 1

icang [17]

Answer:

Approximately $0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}$ .

Explanation:

The specific heat of a material is the amount of energy required to increase unit mass (one gram) of this material by unit temperature (one degree Celsius.)

Calculate the increase in the temperature of this sample:

$\Delta T = (116.9 - 24.3)\; \rm ^\circ\! C= 92.6\; \rm ^\circ\! C$ .

The energy that this sample absorbed should be proportional the increase in its temperature (assuming that no phase change is involved.)

It took $2911\; \rm J$ of energy to raise the temperature of this sample by $\Delta T = 92.6\; \rm ^\circ\! C$ . Therefore, raising the temperature of this sample by $1\; \rm ^\circ\! C$ (unit temperature) would take only $\displaystyle \frac{1}{92.6}$ as much energy. That corresponds to approximately $31.436\; \rm J$ of energy.

On the other hand, the energy required to raise the temperature of this material by $1\; \rm ^\circ\! C$ is proportional to the mass of the sample (also assuming no phase change.)

It took approximately $31.436\; \rm J$ of energy to raise the temperature of $57.07\; \rm g$ of this material by $1\; \rm ^\circ C$ . Therefore, it would take only $\displaystyle \frac{1}{57.07}$ as much energy to raise the temperature of $1\; \rm g$ (unit mass) of this material by $1\; \rm ^\circ \! C\!$ . That corresponds to approximately $0.551\; \rm J$ of energy.

In other words, it takes approximately $0.551\; \rm J$ to raise $1\; \rm g$ (unit mass) of this material by $1\; \rm ^\circ \! C$ . Therefore, by definition, the specific heat of this material would be approximately $0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}$ .

8 0

3 years ago