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3 years ago

How much heat is released as 5.00G of PB cool from 75.0° C to 25.0° C

Chemistry

1 answer:

zalisa [80]3 years ago

7 0

Answer:

32.25 J

Explanation:

Heat, H = ?

Mass, m = 5g = 5 / 1000 = 0.005 Kg

Initial Temperature = 75.0° C

Final Temperature = 25.0° C

Temperature change, ΔT = Final - Initial = 50° C

Specific heat capacity of lead, C = 129 (J/kg°C)

The relationship between these quantities is given by the equation

H = mCΔT

H = 0.005 * 129 * 50

H = 32.25 J

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Answer:

Your correct.

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Ipatiy [6.2K]

Answer:

Reasonable Second step- $N_{2}O(g)+H_{2}(g)\rightarrow N_{2}(g)+H_{2}O(g)$

Explanation:

The given single step chemical reaction is as follows.

$2H_{2}(g)+2NO \rightarrow 2H_{2}O(g)+N_{2}(g)$

Suppose a two-step mechanism is proposed for this reaction,

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$Step -1:H_{2}(g)+2NO \rightarrow N_{2}O(g)+H_{2}O(g)$

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3 years ago

A sheet of gold weighing 8.8 g and at a temperature of 10.5°C is placed flat on a sheet of iron weighing 19.5 g and at a tempera

timofeeve [1]

Answer:

$T = 36.393\,^{\textdegree}C$

Explanation:

The contact between the sheet of gold and the sheet of iron allows a heat transfer until thermal equilibrium is done, which means that both sheets have the same temperature:

$-Q_{iron} = Q_{gold}$

$-(0.008\,kg)\cdot (452\,\frac{J}{kg\cdot ^{\textdegree}C} )\cdot (T-54.4\,^{\textdegree}C) = (0.0195\,kg)\cdot (129\,\frac{J}{kg\cdot ^{\textdegree}C} )\cdot (T-10.5\,^{\textdegree}C)$

$-(3.616\,\frac{J}{^{\textdegree}C})\cdot (T-54.4\,^{\textdegree}C) = (2.515\,\frac{J}{^{\textdegree}C})\cdot (T-10.5^{\textdegree}C)$

$-1.438\cdot (T - 54.4^{\textdegree}C) = T-10.5^{\textdegree}C$

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The final temperature is:

$T = 36.393\,^{\textdegree}C$

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3 years ago

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Ne4ueva [31]

Atoms have no electric charge because the protons and electrons "cancel out" each others charges. Neutrons have no charge. What is the atomic number of an element? The atomic number is the number of protons in the atom's nucleus.

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3 years ago

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Suppose 0.981 g of iron (II) iodide is dissolved in 150. mL of a 35.0 m M aqueous solution of silver nitrate. Calculate the fina

yaroslaw [1]

Answer:

Final molarity of iodide ion C(I-) = 0.0143M

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n = (m(FeI(2)))/(M(FeI(2))

Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol

So n = 0.981/309.85 = 0.0031 mol

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3 years ago