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4 years ago

How much heat is released if 7.15 g cao(s) is added to 152 g of h2o(l)? cao(s) + h2o(l) → ca(oh)2(s) δh°rxn = –64.8 kj/mol?

1 answer:

6 0

First, we calculate the number of moles of each reactant using the formula:

Moles = mass / molecular weight

CaO:
Moles = 7.15/56 = 0.128

Water:
Moles = 152/18 = 8.44

The reaction equation shows that the reactants must be present in an equal number of moles, so CaO will be the limiting reactant and 0.128 mole of calcium hydroxide will form.

The energy released is given by:

Heat of reaction * number of moles
= -64.8 * 0.128
= -8.29 kJ

8.29 kJ of energy will be released

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Is the answer B? Help

Masja [62]

Answer:

Explanation:

Hmm, so we have the following in the diagram

Pt(s)

Cl2(g)

Ag(s)

NaCl(aq)

AgNO3(aq)

Pt 2+, 4+, 6+ Though it states Pt is inert

Cl 2-

Ag 1+

Na 1+

NO3-

Anode definition: the positively charged electrode by which the electrons leave an electrical device.

Electrode definition: a conductor through which electricity enters or leaves an object, substance, or region.

Cations attracted to cathode pick up electrons

Anions attracted to anode release electrodes+

Reduction at Cathode (red cat gain of e)

Oxidation at Anode (ox anode loss of e)

So from the diagram we can see that the charge is being generated through the 2 metal plates.

So the answer is A, the anode material is Pt and the half reaction is 2Cl- = Cl2 + 2e-

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3 years ago

The element iridium exists in nature as two isotopes: 191Ir has a mass of 190.9606 u, and 193Ir has a mass of 192.9629 u. The av

nlexa [21]

Answer: The percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope be 'x'. So, fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope will be '1 - x'

For $_{77}^{191}\textrm{Ir}$ isotope:

Mass of $_{77}^{191}\textrm{Ir}$ isotope = 190.9606 amu

Fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope = x

For $_{77}^{193}\textrm{Ir}$ isotope:

Mass of $_{77}^{193}\textrm{Ir}$ isotope = 192.9629 amu

Fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

$192.22=[(190.9606\times x)+(192.9629\times (1-x))]\\\\x=0.3710$

Percentage abundance of $_{77}^{191}\textrm{Ir}$ isotope = $0.3710\times 100=37.10\%$

Percentage abundance of $_{77}^{193}\textrm{Ir}$ isotope = $(1-0.3710)=0.6290\times 100=62.90\%$

Hence, the percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

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3 years ago

How long does it take for the moon to make one revolution around Earth?

ycow [4]

Answer:

27 days, 7 hours and 43 minutes.

Explanation:

To complete a full orbit around Earth, it takes 27 days, 7 hours and 43 minutes for the Moon. This is called sidereal month and is measured by the Moon's position relative to distant *fixed" stars.

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3 years ago

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lianna [129]

their identified by the number of protons they have in the nucleus which = their electrons.

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4 years ago

What does the author list in the passage?

irina [24]

Answer:

B --> Different types of electromagnetic radiation

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3 years ago

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