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Sunny_sXe [5.5K]

2 years ago

14

certain gas occupies 6.24L at pressure of 760 mm Hg.If the pressure is reduced to 60.0mm Hg,what would the new volume be?

1 answer:

vovikov84 [41]2 years ago

3 0

Answer:

79.04 L

Explanation:

We are given;

Initial Volume; V1 = 6.24L

Initial Pressure; P1 = 760 mm Hg

Final pressure; P2 = 60.0mm Hg

To solve for final volume, we will use Boyles law;

P1•V1 = P2•V2

Let's make V2 which is the final volume the subject;

V2 = (P1•V1)/P2

V2 = (760 × 6.24)/60

V2 = 79.04 L

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You add 50.0 g of ice initially at ‒20.0 °C to 1.00 x 102 mL warm water at 67.0 °C. When all the ice melts, the water temperatur

xxTIMURxx [149]

Answer:

$T_2=17.8\°C$

Explanation:

Hello,

In this case, we can solve this problem by noticing that the heat lost by the warm water is gained by the ice in order to melt it:

$Q_{water}=-Q_{ice}$

In such a way, the cooling of water corresponds to specific heat and the melting of ice to sensible heat and specific heat also that could be represented as follows:

$m_{water}Cp_{water}(T_2-T_{water})=-m_{ice}\Delta H_{melting,ice}-m_{ice}\Cp_{ice}(T_2-T_{ice})$

Thus, specific heat of water is 4.18 J/g°C, heat of melting is 334 J/g and specific heat of ice is 2.04 J/g°C, thus, we can compute the final temperature as shown below:

$m_{water}Cp_{water}(T_2-T_{water})+m_{ice}Cp_{ice}(T_2-T_{ice})=-m_{ice}\Delta H_{melting,ice}\\\\T_2=\frac{-m_{ice}\Delta H_{melting,ice}+m_{water}Cp_{water}T_{water}+m_{ice}Cp_{ice}T_{ice}}{m_{water}Cp_{water}+m_{ice}Cp_{ice}} \\\\T_2=\frac{-50.0*334+100*4.18*67+50.0*2.04*-20.0}{100*4.18+50.0*2.04} \\\\T_2=17.8\°C$

Best regards.

3 0

2 years ago

If you had to explain to a friend how to balance a chemical equation, what

Evgesh-ka [11]

Answer:

make sure that the number of atoms on the left side of the equation equals the number of atoms on the right.

Explanation:

5 0

2 years ago

Zingerone, one of the flavor molecules in ginger,

MissTica

The molecular formula of the compound is C12H15O3 hence the molar mass of the compound is 207 g/mol.

We need to obtain the number of moles of carbon, hydrogen and oxygen in the compound;

Carbon = 24.91 g/44g/mol × 1 mole of carbon = 0.566 moles

Mass of carbon = 0.566 moles × 12 g/mol = 6.792 g

Number of moles of hydrogen = 6.522 g/18 g/mol × 2 moles = 0.725 moles

Mass of hydrogen = 0.725 moles × 1 g/mol = 0.725 g

Mass of oxygen = 10 - (6.792 g + 0.725 g) = 2.483 g

Number of moles of oxygen = 2.483 g/16 g/mol = 0.155 moles

Now we must divide through by the lowest number of moles;

C - 0.566/0.155 H - 0.725/0.155 O - 0.155/0.155

C - 4 H - 5 O - 1

The simplest formula is C4H5O Recall that the molar mass of the compound lies between 150.0 and 220.0 g/mol

4(12) + 5(1) + 16 = 69

Hence; n = 3 and the molecular formula of the compound is C12H15O3

The molar mass of the compound is; 12(12) + 15(1) + 3(16) = 207 g/mol

Learn more: brainly.com/question/15180604

3 0

2 years ago

Which type of bond is present in hydrogen sulfide (H2S)? The table of electronegativities is given.

Paraphin [41]

nonpolar covalent bond (with LD force)

6 0

2 years ago

What is the limiting reactant for the following balanced equation when 9 moles of AlF3 are mixed with 12 miles of O2?

tamaranim1 [39]

<h2>Answer: $AlF_{3}$ </h2>

Explanation:

The chemical equation of the reaction that occurs when $AlF_{3}$ reacts with $O_{2}$ is

$4AlF_{3}+3O_{2}$ → $2Al_{2}O_{3}+6F_{2}$

$4$ moles of $AlF_{3}$ requires $3$ moles of $O_{2}$ .

$1$ mole of $AlF_{3}$ requires $\frac{3}{4}$ moles of $O_{2}$ .

Given that we have $9$ moles of $AlF_{3}$ .

$9$ moles of $AlF_{3}$ requires $\frac{3}{4}\times 9=6.75$ moles of $O_{2}$ .

But we have $12$ moles of $O_{2}$ .

So, $AlF_{3}$ will be consumed first.

So, $AlF_{3}$ is the limiting reagent.

3 0

3 years ago