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3 years ago

3. Burns from boiling water can be severe, caused by the transfer of energy from the boiling water to the

Chemistry

1 answer:

Free_Kalibri [48]3 years ago

8 0

Answer:

Q = 3937.56 J

Explanation:

Heat transferred due to change in temperature is given by :

$Q=mc\Delta T$

c is the specific heat of water, c=4.18 J/g-°C

We have, m = 15 g, $T_i=100^{\circ} C\ \text{and}\ T_f=37.2^{\circ} C$

So,

$Q=15\times 4.18\times (37.2-100)\\Q=-3937.56\ J$

Hence, 3937.56 J of heat is transferred.

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Which of the following statements is true for an exothermic reaction?

shutvik [7]

The enthalpy change for an exothermic reaction is negative because heat is being released, so that takes out two of the responses. Since energy is being released into the surroundings due to the exothermic reaction, the potential energy of the products is lower than that of the reactants. Energy is being put in to make the reaction occur, but then that energy is all being released into the surroundings thus a lower potential energy level for the products

6 0

3 years ago

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How many moles of water can be produced with 4.3 moles of H2 and 5.6 moles of O2? Which reactant is limiting? How many moles of

xeze [42]

Answer:

Hydrogen H₂ will be the limiting reagent.

The excess reactant that will be left after the reaction is 3.45 moles.

4.3 moles of water can be produced.

Explanation:

The balanced reation is:

2 H₂ + O₂ → 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

H₂: 2 moles
O₂: 1 mole
H₂O: 2 moles

To determine the limiting reagent, you can use a simple rule of three as follows: if by stoichiometry 1 mole of O₂ reacts with 2 moles of H₂, how much moles of H₂ will be needed if 5.6 moles of O₂ react?

$moles of H_{2} =\frac{5.6 moles of O_{2} *2 mole of H_{2} }{1 mole of O_{2}}$

moles of H₂= 11.2 moles

But 11.2 moles of H₂ are not available, 4.3 moles are available. Since you have less moles than you need to react with 5.6 moles of O₂, hydrogen H₂ will be the limiting reagent and oxygen O₂ will be the excess reagent.

Then you can apply the following rules of three:

If by reaction stoichiometry 2 moles of H₂ react with 1 mole of O₂, 4.3 moles of H₂ will react with how many moles of O₂?

$moles of O_{2} =\frac{1 mole of O_{2} *4.3 mole of H_{2} }{2 mole of O_{2}}$

moles of O₂= 2.15 moles

The excess reactant that will be left after the reaction can be calculated as:

5.6 moles - 2.15 moles= 3.45 moles

The excess reactant that will be left after the reaction is 3.45 moles.

If by reaction stoichiometry 2 moles of H₂ produce 2 moles of H₂O, 4.3 moles of H₂ produce how many moles of H₂O?

$moles of H_{2}O =\frac{2 moles of H_{2}O *4.3 mole of H_{2} }{2 mole of H_{2}}$

moles of H₂O= 4.3 moles

4.3 moles of water can be produced.

8 0

2 years ago

When a sodium atom becomes an ion the size of the atom?

jeyben [28]

The size of atom reduces

Electronic configuration of sodium -
2,8,1

In order to become stable it loses one electron and becomes positive ion.
It's size reduces.

3 0

3 years ago

the value of ksp for pbcl2 is 1.6. what is the lowest concentration of Cl- that would be needed to begin precipitation of PbCl2

deff fn [24]

Answer:

The minimum concentration of Cl⁻ that produces precipitation is 12.6M

Explanation:

The Ksp of PbCl₂ is expressed as:

PbCl₂(s) → Pb²⁺(aq) + 2Cl⁻(aq)

The Ksp is:

Ksp = 1.6 = [Pb²⁺] [Cl⁻]²

When Ksp = [Pb²⁺] [Cl⁻]² the solution begind precipiration.

A 0.010M Pb(NO₃)₂ is 0.010M Pb²⁺, thus:

1.6 = [0.010M] [Cl⁻]²

160 = [Cl⁻]²

12.6M = [Cl⁻]

<h3>The minimum concentration of Cl⁻ that produces precipitation is 12.6M</h3>

7 0

2 years ago

Which branch of chemistry specifically deals with quantitative analysis?

LUCKY_DIMON [66]

Answer:

analyitical

Explanation:

4 0

3 years ago

Read 2 more answers