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disa [49]
2 years ago
12

A piece of iron is heated to 120'C then placed in a calorimeter containing 50g of water. The water temperature is raised from 30

°C to 40°C. Assuming the specific heat of water is 4. 18 J/8'C and the specific heat of
Iron is 0. 44 J/g C, what is the mass of the piece of iron?

118. 8

59. 45

39. 6

752
Chemistry
1 answer:
Lana71 [14]2 years ago
3 0

Answer:

Mass of iron = 59.375 gm

Explanation:

Calories ( or joules) are added to the water by the hot steel so at the endpoint they are BOTH at 40 C

 The water gains:

     4.18  j/g-C  * 50 * (40-30 C) = 2090 j

The steel gave up 2090 j going from 120 to 40 C

2090 = .44 j/g-C  * m * (120-40)    solve fro m = 59.375 gm

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increases as you move down a group as the number of electrons increases. Therefore, the atomic radius increases as the group and energy levels increase

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3 years ago
if a gas has an initial pressure of 24,650 pa and an initial volume of 376 ml, what is the final volume if the pressure of the g
NNADVOKAT [17]

If a gas has an initial pressure of 24,650 pa and an initial volume of 376 ml, then the final volume would be 11,943.8144 ml if the pressure of the gas is changed to 775 torr assuming that the amount and the temperature of the gas remain constant.

It is given that the initial pressure P₁ is 24,650Pa and initial volumeV₁ is 376ml and the final pressureP₂ is 775 torr. We need to find the final volume of the gas. The final volume could be found using the following formula:

P₁V₁ = P₂V₂

By substituting the values, we get

24650 x 376 = 776 x V₂

9268400 = 776V₂

V₂ = 9268400/776

V₂ = 11,943.8144 ml

Therefore, the final volume of the gas would be 11,943.8144 ml

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5 0
1 year ago
Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp
Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}

For calculating edge length,

a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}

For calculating M_{ave}, we use the formula

M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}

Similarly for calculating (\rho)_{ave}, we use the formula

\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}

From the periodic table, masses of the two elements can be written

M_{Fe}= 55.85g/mol

M_{V}=50.941g/mol

Specific density of both the elements are

(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3

Putting  M_{ave} and \rho_{ave} formula's in edge length formula, we get

a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}}  \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}}  \right )}  \right ]^{1/3}

a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}}  \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}}  \right )}  \right ]^{1/3}

By calculating, we get

a=2.89\times10^{-8}cm=0.289nm

7 0
3 years ago
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