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torisob [31]
3 years ago
8

15. An increase in atomic number is related to an increase in atomic mass because

Chemistry
1 answer:
NARA [144]3 years ago
6 0

Answer:

\boxed {\boxed {\sf C. \ More \ protons \ are \ present \ in \ the \ atomic \ nucleus}}

Explanation:

First, let's define <u>atomic number</u>.

    ⇒ The number of <u>protons </u>in an atom.

    ⇒ This number determines the identity and properties of the atom

So, atomic number relates to protons. There is an increase in atomic number, so there must be <u>more</u> protons in the atom. We can eliminate choices A and B, because they pertain to electrons.

We are left with choices C and D. The difference between them is the location of the protons.

Protons are part of the <u>nucleus </u>of the atom: the dense, central region of the atom comprised of nucleons: protons and neutrons.

Electrons, not protons, orbit the nucleus, so  D is not correct.

We are left with choice C: <u>more protons are present in the atomic nucleus </u>

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3 years ago
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How much heat do you need to raise the temperature of 100g of aluminum from 30 C to 150 C
SVEN [57.7K]

Answer:

Q = 10.8 KJ

Explanation:

Given data:

Mass of Al= 100g

Initial temperature = 30°C

Final temperature = 150°C

Heat required = ?

Solution:

Specific heat of Al = 0.90 j/g.°C

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 150°C - 30°C

ΔT = 120°C

Q = 100g×0.90 J/g.°C× 120°C

Q = 10800 J       (10800j×1KJ/1000 j)

Q = 10.8 KJ

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maxonik [38]

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3 years ago
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MULTIPLE CHOICE
adelina 88 [10]

Answer:

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3 years ago
What volume of benzene (C6H6, d= 0.88 g/mL, molar mass = 78.11 g/mol) is required to produce 1.5 x 103 kJ of heat according to t
Anton [14]

Answer:

We need 42.4 mL of benzene to produce 1.5 *10³ kJ of heat

Explanation:

<u>Step 1:</u> Data given

Density of benzene = 0.88 g/mL

Molar mass of benzene = 78.11 g/mol

Heat produced = 1.5 * 10³ kJ

<u>Step 2:</u> The balanced equation

2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g)         ΔH°rxn = -6278 kJ

<u>Step 3:</u> Calculate moles of benzen

1.5 * 10³ kJ * (2 mol C6H6 / 6278 kJ) = 0.478 mol C6H6

<u>Step 4:</u> Calculate mass of benzene

Mass benzene : moles benzene * molar mass benzene

Mass benzene= 0.478 * 78.11 g

Mass of benzene = 37.34 grams

<u>Step 5:</u> Calculate volume of benzene

Volume benzene = mass / density

Volume benzene = 37.34 grams / 0.88g/mL

Volume benzene = 42.4 mL

We need 42.4 mL of benzene to produce 1.5 *10³ kJ of heat

4 0
3 years ago
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