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3 years ago

Is anyone good at 10th grade chemistry ? If so can you help me please

Chemistry

1 answer:

Harrizon [31]3 years ago

5 0

Answer:

221.7 K

Explanation:

As, PV = nRT

Hence,

6.5 (14) = 5(0.0821)T

91 = 0.4105 T

91 ÷ 0.4105 = T

221.7 = T

Hence the required temperature is 221.7 K (Ans)

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A flask containing helium gas is connected to an open-ended mercury manometer. The open end is exposed to the atmosphere, where

stepladder [879]

Answer:

726 torr

Explanation:

Generally, atmospheric pressure can be measured using a manometer which is in form of a U-shaped tube. In addition, 1 mm Hg is equivalent to 1 torr. Therefore, 752 torr is equivalent to 752 mm Hg. Therefore, the total pressure will be equivalent to the atmospheric pressure (mm Hg) + the mercury height.

In this case, the mercury height = -26 mm

Thus:

The helium pressure = 752 - 26 = 726 mm Hg

This is also equivalent to 726 torr

8 0

3 years ago

A drought is a long period of dryness. If animals fight during a drought, which basic need is most likely limited?

kicyunya [14]

Answer:

Water. No water falls from the sky during such times.

7 0

3 years ago

The potential in an electrochemical cell, E, is related to the Gibb's free energy change (ΔG) for the overall cell redox reactio

Nana76 [90]

Answer:

Explanation:

As an example, the following cell reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m) generates a cell voltage of +1.10 V under standard conditions. Calculate and enter delta G degree (with 3 sig figs) for this reaction in kJ/mol.

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(m)

ΔG = ΔG° + RTInQ

Q = 1

ΔG = ΔG°

ΔG = =nFE°

n=no of electrons transfered.

E° = 1.1v

ΔG° = -2 * 96500 * 1.10

= -212300J

ΔG° =-212.3kJ/mol

<h3>Therefore, the ΔG° = -212.3kJ/mol</h3>

5 0

3 years ago

PLEASE PLEASE HELP!

valentinak56 [21]

Answer: The number of grams of $H_2$ in 1620 mL is 1.44 g

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 1 atm (at STP)

V = Volume of gas = 1620 ml = 1.62 L (1L=1000ml)

n = number of moles = ?

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $273K$

$n=\frac{PV}{RT}$

$n=\frac{1atm\times 16.2L}{0.0821Latm/K mol\times 273K}=0.72moles$

Mass of hydrogen = $moles\times {\text {Molar mass}}=0.72mol\times 2g/mol=1.44g$

The number of grams of $H_2$ in 1620 mL is 1.44 g

8 0

3 years ago

Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron. A mixtur

yuradex [85]

Answer: The limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For magnesium:

Given mass of magnesium = 41.0 g

Molar mass of magnesium = 24 g/mol

Putting values in equation 1, we get:

$\text{Moles of magnesium}=\frac{41.0g}{24g/mol}=1.708mol$

For iron(III) chloride:

Given mass of iron(III) chloride = 175.0 g

Molar mass of iron(III) chloride = 162.2 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron(III) chloride}=\frac{175g}{162.2g/mol}=1.708mol$

The chemical equation for the reaction of magnesium and iron(III) chloride follows:

$3Mg+2FeCl_3\rightarrow 3MgCl_2+2Fe$

By Stoichiometry of the reaction:

3 moles of magnesium reacts with 2 moles of iron(III) chloride

So, 1.708 moles of magnesium will react with = $\frac{2}{3}\times 1.708=1.114mol$ of iron(III) chloride

As, given amount of iron(III) chloride is more than the required amount. So, it is considered as an excess reagent.

Thus, magnesium is considered as a limiting reagent because it limits the formation of product.

Moles of excess reactant left (iron(III) chloride) = [1.708 - 1.114] = 0.594 moles

Now, calculating the mass of iron(III) chloride from equation 1, we get:

Molar mass of iron(III) chloride = 162.2 g/mol

Moles of iron(III) chloride = 0.594 moles

Putting values in equation 1, we get:

$0.594mol=\frac{\text{Mass of iron(III) chloride}}{162.2g/mol}\\\\\text{Mass of iron(III) chloride}=(0.594mol\times 162.2g/mol)=96.35g$

Hence, the limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.

6 0

3 years ago