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babymother [125]
3 years ago
14

It takes 5.0 J of heat in increase the temp of a sample by 10 degrees Celcius. If the mass is doubled, how much heat in joules w

ill it take to increase the temp by 10.0 degrees.a. 50.0 Jb. 5.0 Jc. 10.0 Jd. 15.0 J
Chemistry
1 answer:
Alona [7]3 years ago
5 0

Answer:

c. 10.0 J

Explanation:

The heat (Q) needed to increase the temperature of a certain mass of a sample can be calculated using the following expression:

Q = c . m . ΔT

where,

c is the heat capacity

m is the mass

ΔT is the change in temperature

Let's call Q₁ (5.0 J) the heat to increase the temperature of the mass m₁ by ΔT = 10.0°C.

Q₁ = c . m₁ . ΔT

If the mass is doubled (m₂ = 2 m₁), the heat required is:

Q₂ = c . m₂ . ΔT = c . 2 . m₁ . ΔT = 2 . c . m₁ . ΔT = 2 . Q₁

Q₂ = 2 . Q₁ = 2 . 5.0 J = 10.0 J

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Explanation:

Octet, in chemistry, the eight-electron arrangement in the outer electron shell of the noble-gas atoms. This structure is held responsible for the relative inertness of the noble gases and the chemical behaviour of certain other elements.

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describe the trends in the atomic size, location energy and electrongavity from left to right across a period in periodic table
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Atomic size decreases in a period but the ionization energy and electronegativity increases across a period.

<h3>Describe the trends in the atomic size, ionization energy and electronegativity?</h3>

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Ionization energy increases when we move from left to right across an period and decreases from top to bottom.

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3 0
2 years ago
One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate
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Answer:

\delta H_{rxn} = -66.0  \ kJ/mole

Explanation:

Given that:

3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \  \ \delta H = -47.0 \ kJ/mole  -- equation (1)  \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)}  \ \ \delta H = -25.0 \ kJ/mole  -- equation (2)  \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole  -- equation (3)

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

3FeO_{(s)} + CO_{2(g)}    \to    Fe_3O_4_{(s)} + CO_{(g)}   \ \delta H = -19.0 \ kJ/mole  -- equation (4)

Multiplying  (2) with equation (4) ; we have:

6FeO_{(s)} + 2CO_{2(g)}    \to    2Fe_3O_4_{(s)} + 2CO_{(g)}   \ \delta H = -38.0 \ kJ/mole  -- equation (5)

From equation (1) ; multiplying (-1) with equation (1); we have:

2Fe_3O_4_{(s)} +CO_{2(g)} \to     3FeO_3_{(s)}+CO_{(g)}   \  \ \delta H = 47.0 \ kJ/mole  -- equation (6)

From equation (2); multiplying (3) with equation (2); we have:

3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)}  \ \ \delta H = -75.0 \ kJ/mole  -- equation (7)

Now; Adding up equation (5), (6) & (7) ; we get:

6FeO_{(s)} + 2CO_{2(g)}    \to    2Fe_3O_4_{(s)} + 2CO_{(g)}   \ \delta H = -38.0 \ kJ/mole  -- equation (5)

2Fe_3O_4_{(s)} +CO_{2(g)} \to     3FeO_3_{(s)}+CO_{(g)}   \  \ \delta H = 47.0 \ kJ/mole  -- equation (6)

3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)}  \ \ \delta H = -75.0 \ kJ/mole  -- equation (7)

<u>                                                                                                                      </u>

FeO  \ \ \ +  \ \ \ CO   \ \  \to   \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \  \delta H = - 66.0 \ kJ/mole

<u>                                                                                                                     </u>

<u />

\delta H_{rxn} = \delta H_1 +  \delta H_2 +  \delta H_3    (According to Hess Law)

\delta H_{rxn} = (-38.0 +  47.0 + (-75.0)) \ kJ/mole

\delta H_{rxn} = -66.0  \ kJ/mole

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