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3 years ago

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A hot metal plate at 150°C has been placed in air at room temperature. Which event would most likely take place over the next fe

w minutes? Molecules in both the metal and the surrounding air will start moving at lower speeds. Molecules in both the metal and the surrounding air will start moving at higher speeds. The air molecules that are surrounding the metal will slow down, and the molecules in the metal will speed up. The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down. Mark this and return

1 answer:

spayn [35]3 years ago

5 0

Cyndi hang Geld Robin

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A certain liquid has a normal freezing point of and a freezing point depression constant . Calculate the freezing point of a sol

katrin2010 [14]

The question is incomplete, the complete question is:

A certain liquid X has a normal freezing point of $0.80^oC$ and a freezing point depression constant $K_f=7.82^oC.kg/mol$ . Calculate the freezing point of a solution made of 81.1 g of iron(III) chloride () dissolved in 850. g of X. Round your answer to significant digits.

<u>Answer:</u> The freezing point of the solution is $-17.6^oC$

<u>Explanation:</u>

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m$

OR

$\text{Freezing point of pure solvent}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

where,

Freezing point of pure solvent = $0.80^oC$

Freezing point of solution = $?^oC$

i = Vant Hoff factor = 4 (for iron (III) chloride as 4 ions are produced in the reaction)

$K_f$ = freezing point depression constant = $7.82^oC/m$

$m_{solute}$ = Given mass of solute (iron (III) chloride) = 81.1 g

$M_{solute}$ = Molar mass of solute (iron (III) chloride) = 162.2 g/mol

$w_{solvent}$ = Mass of solvent (X) = 850. g

Putting values in equation 1, we get:

$0.8-(\text{Freezing point of solution})=4\times 7.82\times \frac{81.1\times 1000}{162.2\times 850}\\\\\text{Freezing point of solution}=[0.8-18.4]^oC\\\\\text{Freezing point of solution}=-17.6^oC$

Hence, the freezing point of the solution is $-17.6^oC$

6 0

2 years ago

When you combine 50.0 mL of 0.100 M AgNO3 with 50.0 mL of 0.100 M HCl in a coffee-cup calorimeter, the temperature changes from

RideAnS [48]

Answer : The enthalpy of reaction $(\Delta H_{rxn})$ is, 67.716 KJ/mole

Explanation :

First we have to calculate the moles of $AgNO_3$ and $HCl$ .

$\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole$

$\text{Moles of }HCl=\text{Molarity of }HCl\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole$

Now we have to calculate the moles of AgCl formed.

The balanced chemical reaction will be,

$AgNO_3(aq)+HCl(aq)\rightarrow AgCl(s)+HNO_3(aq)$

As, 1 mole of $AgNO_3$ react with 1 mole of $HCl$ to give 1 mole of $AgCl$

So, 0.005 mole of $AgNO_3$ react with 0.005 mole of $HCl$ to give 1 mole of $AgCl$

The moles of AgCl formed = 0.005 mole

Total volume of the solution = 50.0 ml + 50.0 ml = 100.0 ml

Now we have to calculate the mass of solution.

Mass of the solution = Density of the solution × Volume of the solution

Mass of the solution = 1.00 g/ml × 100.0 ml = 100 g

Now we have to calculate the heat.

$q=m\times C\Delta T=m\times C \times (T_2-T_1)$

where,

q = heat

C = specific heat capacity = $4.18J/g^oC$

m = mass = 100 g

$T_2$ = final temperature = $24.21^oC$

$T_1$ = initial temperature = $23.40^oC$

Now put all the given values in the above expression, we get:

$q=100g\times (4.18J/g^oC)\times (24.21-23.40)^oC$

$q=338.58J$

Now we have to calculate the enthalpy of the reaction.

$\Delta H_{rxn}=\frac{q}{n}$

where,

$\Delta H_{rxn}$ = enthalpy of reaction = ?

q = heat of reaction = 338.58 J

n = moles of reaction = 0.005 mole

Now put all the given values in above expression, we get:

$\Delta H_{rxn}=\frac{338.58J}{0.005mole}=6771.6J/mole=67.716KJ/mole$

Conversion used : (1 KJ = 1000 J)

Therefore, the enthalpy of reaction $(\Delta H_{rxn})$ is, 67.716 KJ/mole

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3 years ago

Which of four fossils would be the most useful as a reference to determine the age of newly discovered fossils?why?

weqwewe [10]

Answer:

(A) fossil is the most useful.

It serves as an index fossil. The seashell is a great index fossil because it only appears in the sandstone layer in multiple locations in the same layer of time.

Explanation:

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2 years ago

How do polar and nonpolar covalent bonds differ?

3241004551 [841]

Answer: In polar covalent bonds a pair of electrons are unequally shared between two atoms, while in nonpolar covalent bonds two atoms share a pair of electrons with each other.

I hope this helps :)

Explanation:

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3 years ago

How many hydrogen bonds are between cytosine and guanine

Dahasolnce [82]

Answer:

3

Explanation:

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2 years ago

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