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-BARSIC- [3]
3 years ago
7

If the temperature of 15 grams of water changes from 21C to 24C, how many joules of heat were involved? Show work

Chemistry
1 answer:
goblinko [34]3 years ago
6 0

Answer:

189 Joules

Explanation:

Applying,

Q = cm(t₂-t₁)............. equation 1

Where Q = Heat, c = specific heat capacity of water, m = mass of water, t₁ = Initial Temperature, t₂ = Final temperature.

From the question,

Given: m = 15 grams = 0.015 kg, t₁ = 21 °C, t₂ = 24 °C

Constant: c = 4200J/kg.°C

Substitute these values into equation 1

Q = 0.015×4200×(24-21)

Q = 0.015×4200×3

Q = 189 Joules

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When a new bond forms in a chemical reaction, what happens?
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Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
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Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, C_{Ag} = 78 wt%

concentration of Pd, C_P_d = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, A_A_g = 107.87 g/mol

atomic weight of iron, A_P_d = 106.4 g/mol

Step 1: determine the average density of the alloy

\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }

\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3

Step 2: determine the average atomic weight of the alloy

A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }

A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole

Step 3: determine unit cell volume

V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell

Step 4: determine the unit cell edge length

Vc = a³

a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm= 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0
3 years ago
Please someone can help with this answer?
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Answer:

6,000kg/m3

Explanation:

6.00g/1cm3 x 1kg/1000g x 1cm3/0.000001m3

= 6.00kg/0.001m3

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5 0
3 years ago
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