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erastovalidia [21]
3 years ago
14

A stack of 10 pennies weighing 26.55 grams is placed in a graduated cylinder containing water. The initial volume of water is 50

.0 mL and the final volume of the water and pennies 53.3 mL. What is the density of ONE penny?
Chemistry
1 answer:
BigorU [14]3 years ago
7 0
In order to determine the density of an item, we will need to determine its mass and volume. The standard unit for measuring mass in a lab is the gram. Think about liquids- what units do you typically report the volume of a liquid in? What about for a sugar cube, what volume is the most appropriate?
A regular object like a sugar cube can be measured with a ruler so we might report the volume in centimeters cubed (cm3). An irregular object like the plate pictured below can be measured by using a technique called volume by displacement. A liquid (typically water) is placed in a graduated cylinder and the volume of a liquid is measured. Then the irregular object is placed in the liquid and the volume is measured again. The change in volume is the irregular object’s volume. This measurement is often made using a graduated cylinder and recording a volume in Liters or milliliters (mL).
Figure 1. (a) Regular object of metal blocks with the same width, length, and height. (B) An irregular
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Why would you need to heat the hydrate up to 250 degrees if water boils at 100 degrees Celsius? Do you think that the boiling po
sergeinik [125]

Explanation:

The pure form of water has a boiling point of 100°C. Boiling point is a physical property of matter and it shows that at such temperature, a liquid substance will change state to vapor.

Pure water is made up of 2 atoms of hydrogen and 1 atom of oxygen. The only intermolecular forces between them is the hydrogen bonds that must be broken for the water to boiling off.

In hydrate, water is present in another form. The water is attached to another compound.

For a pure liquid, the they have reasonably constant boiling point and low boiling range.

Impurities such as the other part of the hydrate causes the elevation of the boiling point and the widening of the boiling range for impure substances.

We are no longer dealing with just hydrogen bonds, other molecular interactions are now involved and they need to be accounted for.

learn more:

Pure substances brainly.com/question/1832352

#learnwithBrainly

8 0
3 years ago
Emission Spectrum Questions for Quiz Print - Quizizz
-Dominant- [34]

Isotopes of  the same element

₃₅⁷⁷X and ₃₅⁸¹X

<h3>Further explanation</h3>

Given

Isotopes of element

Required

Isotopes of  the same element

Solution

The elements in nature have several types of isotopes  

Isotopes are elements that have the same Atomic Number (Proton)  and different mass numbers

Element symbols that meet the 2 conditions above are :

₃₅⁷⁷X and ₃₅⁸¹X

3 0
2 years ago
A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %
kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and y_{C-6} and y_{N_2} are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

Q=-40036+(-672.21)=-40708.21J

Finally we convert this result to kJ:

Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ

Learn more:

  • brainly.com/question/25475410
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6 0
2 years ago
Diatomic iodine [I2] decomposes at high temperature to form I atoms according to the reaction I2(g)⇌2I(g), Kc = 0.011 at 1200∘C
umka2103 [35]

Answer:

The concentration of I at equilibrium = 3.3166×10⁻² M

Explanation:

For the equilibrium reaction,

I₂ (g) ⇄ 2I (g)

The expression for Kc for the reaction is:

K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}

Given:

\left[I_2_{Equilibrium} \right] = 0.10 M

Kc = 0.011

Applying in the above formula to find the equilibrium concentration of I as:

0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}

So,

\left[I_{Equilibrium} \right]^2=0.011\times 0.10

\left[I_{Equilibrium} \right]^2=0.0011

\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M

<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>

3 0
3 years ago
Aluminum reacts with chlorine gas to form aluminum chloride via the following reaction:
Leni [432]

Answer:

science ka Hindi kya hoga

8 0
2 years ago
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