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3 years ago

A 63.5 g sample of an unidentified metal absorbs 355 ) of heat when its temperature changes

Chemistry

1 answer:

insens350 [35]3 years ago

5 0

0.208 is the specific heat capacity of the metal.

Explanation:

Given:

mass (m) = 63.5 grams 0R 0.0635 kg

Heat absorbed (q) = 355 Joules

Δ T (change in temperature) = 4.56 degrees or 273.15+4.56 = 268.59 K

cp (specific heat capacity) = ?

the formula used for heat absorbed and to calculate specific heat capacity of a substance will be calculated by using the equation:

q = mc Δ T

c = $\frac{q}{mΔ T}$

c = $\frac{355}{63.5X 268.59}$

= 0.208 J/gm K

specific heat capacity of 0.208 J/gm K

The specific heat capacity is defined as the heat required to raise the temperature of a substance which is 1 gram. The temperature is in Kelvin and energy required is in joules.

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1. A dining hall had a total of 25 tables-some long rectangular tables and some round

TEA [102]

x = 20 long tables

y = 5 round table

Explanation:

We have the following system of equations:

x + y = 25

8x + 6y = 190

From the first equation we have:

x = 25 - y

And we replace x in the second equation:

8(25 - y) + 6y = 190

200 - 8y + 6y = 190

200 - 2y = 190

200 - 190 = 2y

10 = 2y

y = 5

Now we insert the value of y in the next equation:

x = 25 - y

x = 25 - 5

x = 20

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3 years ago

Coal power plants burn large amounts of coal, c(s), in an o 2 (g) atmosphere to generate electricity. the chemical reaction re

charle [14.2K]

Answer is: volume of carbon dioxide is 1,84·10⁸ l.
Chemical reaction: C + O₂ → CO₂.
m(C) = 100 t · 1000 kg/t = 100000 kg
m(C) = 100000 kg · 1000 g/kg = 10⁸ g.
n(C) = m(C) ÷ M(C).
n(C) = 10⁸ g ÷ 12 g/mol.
n(C) = 8,33·10⁶ mol.
From chemical reaction: n(C) . n(CO₂) = 1 : 1.
n(CO₂) = 8,33·10⁶ mol.
m(CO₂) = 8,33·10⁶ mol · 44 g/mol.
m(CO₂) = 3,66·10⁸ = 3,66·10⁵ kg.
V(CO₂) = 3,66·10⁵ kg ÷ 1,98 kg/m³ = 1,84·10⁵ m³.
V(CO₂) = 1,84·10⁵ m³ · 1000 l/m³ = 1,84·10⁸ l.

6 0

3 years ago

The half-life for the radioactive decay of C-14C-14 is 5730 years. You may want to reference (Pages 598 - 605) Section 14.5 whil

KIM [24]

<u>Answer:</u> The sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

$t_{1/2}=\frac{0.693}{k}$

where,

$t_{1/2}$ = half life of the reaction = 5730 years

Putting values in above equation, we get:

$k=\frac{0.693}{5730yrs}=1.21\times 10^{-4}yrs^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$

where,

k = rate constant = $1.21\times 10^{-4}yr^{-1}$

t = time taken for decay process = ? yr

$[A_o]$ = initial amount of the sample = 100 grams

[A] = amount left after decay process = (100 - 25) = 75 grams

Putting values in above equation, we get:

$1.21\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{75}\\\\t=2377.9yrs$

Hence, the sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

8 0

3 years ago

51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b

Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

$a = \sqrt{8} r$

$a = \sqrt{8} \times 144 pm$

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}$

density d = 19.41 g/cm³

Similarly; For a body-centered cubic structure

$r = \dfrac{\sqrt{3}}{4}a$

where;

r = 144

$144 = \dfrac{\sqrt{3}}{4}a$

$a = \dfrac{144 \times 4}{\sqrt{3}}$

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V 3.68 × 10⁻²³ cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

$density = \dfrac{mass}{volume}$

$density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}$

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0

3 years ago

You have collected a tissue specimen that you would like to preserve by freeze drying. To ensure the integrity of the specimen,

Reika [66]

Answer:

The maximum pressure is 612.2 Pa

Explanation:

The pressure of the ice (P1) = 624 Pa

The temperature of the ice = 273.16 K

The maximum temperature the specimen = - 5 oC

= -5 + 273 = 268K

The maximum Pressure the freeze drying can be will be (P2) = ?

Using Pressure law, which shows the relationship between pressure and temperature.

P1 / T1 = P2 / T2

P2 T1 = P1 T2

P2 = P1 T2 / T1

P2 = 624 × 268 / 273.16

P2 = 612.2 Pa

The maximum pressure at which drying can be carried out is 612.2 Pa

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5 0

3 years ago