The area that includes everything between the cell membrane and the nucleus of a cell is called the __________

What is the density of Iodine gas at 2.36 atm and 65.0 C?

timurjin [86]

The density : 21.63 g/L

<h3>Further explanation</h3>

The gas equation can be written

$\large {\boxed {\bold {PV = nRT}}}$

where

P = pressure, atm

V = volume, liter

n = number of moles

R = gas constant = 0.08206 L.atm / mol K

T = temperature, Kelvin

T = 65+273=338 K

P=2.36 atm

MW Iodine gas (I₂)=254

The density :

$\tt \rho=\dfrac{P\times MW}{RT}\\\\\rho=\dfrac{2.36\times 254}{0.082\times 338}=21.63~g/L$

4 0

3 years ago

If I have 4.5 liters of gas at a temperature of 33 0C and a pressure of 6.54 atm, what will be the pressure of the gas if I rais

Nikitich [7]

This is an exercise in the general or combined gas law.

To start solving this exercise, we must obtain the following data:

V₁ = 4.5 l
T₁ = 33 °C + 273 = 306 k
P₁ = 6.54 atm
T₂ = 94 °C + 273 = 367 k
V₂ = 2.3 l
P₂ = ¿?

We use the following formula:

P₁V₁T₂ = P₂V₂T₁ ⇒ General Formula

Where

P₁ = Initial pressure
V₁ = Initial volume
T₂ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₁ = Initial temperature

We clear the general formula for the final pressure.

$\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{P_{1}V_{1}T_{2} }{V_{2}T_{1}} \ \to \ Clear \ formula \end{gathered}$}$

We solve by substituting our data in the formula:

$\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{(6.54 \ atm)(4.5 \not{l})(367 \not{K}) }{(2.3 \not{l})(306 \not{k})} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{10800.81}{ 703.8 } \ atm \end{gathered}$}$

$\boxed{\large\displaystyle\text{$\begin{gathered}\sf P_{2}=15.346 \ atm \end{gathered}$} }$

If I raise the temperature to 94°C and decrease the volume to 2.3 liters, the pressure of the gas will be 15,346 atm.

8 0

2 years ago

Read 2 more answers

The formula of nitrogen oxide is NO, of nitrogen dioxide is NO_2. Write a balanced equation for the reaction of nitrogen oxide w

denis23 [38]

Answer:

a) 0,5 mol O₂; 1 mol NO₂

b)

Liters of NO = 22,4 L

Liters of O₂ = 11,2 L

Liters of NO₂ = 22,4 L

c) NO = 30g

O₂ = 16g

NO₂ = 46g

d) 7,41g HCl

e) 107,7 g/mol

f) 0,0312 moles of O₂; 0,999g of O₂; 699 mL at STP or 803 mL

ii. 51%

g) 32,6L

Explanation:

a) For the reaction:

2 NO + O₂ → 2 NO₂

For 1 mole of NO there are consumed:

1 mol NO × $\frac{1molO_{2}}{2molNO}$ = 0,5 mol of O₂

And produced:

1 mol NO × $\frac{2molNO_{2}}{2molNO}$ = 1 mol of NO₂

b) By ideal gas law:

V = nRT/P

Where n is moles of each compound; R is gas constant (0,082atmL/molK); T is temperature (273,15 K at state conditions); P is pressure (1 atm at STP) and V is volume in liters. Replacing each moles for each compound:

Liters of NO = 22,4 L

Liters of O₂ = 11,2 L

Liters of NO₂ = 22,4 L

c) The mass of each compound are:

1 mol NO× $\frac{30 g}{1molNO}$ = 30g

0,5 mol O₂× $\frac{32 g}{1molO_{2}}$ = 16g

1 mol NO₂× $\frac{46 g}{1molNO_{2}}$ = 46g

d) Using:

n = PV / RT

Moles of 4,55 L of HCl (using the values of P = 1 atm; R = 0,082atmL/molK; T = 273,15K) are:

0,203 moles of HCl. In grams:

0,203 mol HCl× $\frac{36,46 g}{1molHCl}$ = 7,41 g of HCl

e) Using:

δRT/P = MW

Where δ is density in g/L (4,81 g/L); R is gas constant (0,082atmL/molK); T is temperature (273,15K); P is pressure (1 atm)

And MW is molecular mass: 107,7 g/mol

f) For the reaction:

2 KClO₃ → 2 KCl + 3 O₂

2,550 g of KClO₃ are:

2,550 g of KClO₃× $\frac{1mol}{122,55 gKClO_{3}}$ = 0,0208 moles of KClO₃

When these moles reacts completely produce:

0,0208 moles of KClO₃× $\frac{3 mol O_{2}}{2 molKClO_{3}}$ = 0,0312 moles of O₂

In grams:

0,0312 moles of O₂ × $\frac{32g}{1 molO_{2}}$ = 0,999g of O₂

V = nRT/P

At STP, n = 0,0312 mol; R = 0,082atmL/molK;T= 273,15K; P = 1atm; V = 0,699L ≡ 699mL

At 29 °C (302,15K) and 732 torr (0,963 atm)

V = 0,803L ≡ 803mL

ii. 182 mL ≡ 0,182L of O₂ are:

n = PV/RT

moles of O₂ are 7,07x10⁻³. Moles of KClO₃ are:

7,07x10⁻³ moles of O₂× $\frac{2 mol KClO_{3}}{3 molO_{2}}$ = 0,0106 mol KClO₃. In grams:

0,0106 moles of KClO₃× $\frac{122,55 g}{1molKClO_{3}}$ = 1,300 g of KClO₃.

Thus, percent by mass of KClO₃ in the mixture is:

1,300g/2,550g ×100 = 51%

g. Combined gas law says that:

$\frac{P_{1}V_{1}}{T_{1}} =\frac{P_{2}V_{2}}{T_{2}}$

Where:

P₁ = 755 torr; V₁ = 35,9L; T₁ = 26°C (299,15 K); P₂ = 760 torr (STP): T₂ = 273,15K (STP) V₂ = 32,6 L

I hope it helps!

4 0

3 years ago

Name the compounds( Please answer only if you know)

IRISSAK [1]

Answer

2-methyl-2-pentene

Explanation:

1. Identify the group that takes precedence in this case alkene hence this molecule is an alkene with a methyl group side chain.

2.Find the longest carbon chain where the functional group(alkene group in this case) has the lowest Carbon number

3.What are the side groups? One side group can be seen at carbon 2 this group is methyl

4. Naming, number separated by "," and number from letters by "-" so the compound should be

2-methyl-2-pentene

8 0

3 years ago

Calculate the maximum mass of potassium sulfate, K2SO4, that could be formed when 4.9 g of sulfuric acid reacts with excess pota

DerKrebs [107]

Answer:

From 4.9 g of sulfuric acid 8.175 g of potassium sulfate is formed.

Explanation:

Given data:

Mass of H₂SO₄ = 4.9 g

Mass of K₂SO₄ formed = ?

Solution:

Chemical reaction:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

Number of moles of H₂SO₄:

Number of moles = mass/ molar mass

Number of moles = 4.9 g/ 98.1 g/mol

Number of moles = 0.05 mol

Now we will compare the moles of H₂SO₄ with K₂SO₄ .

H₂SO₄ : K₂SO₄

1 : 1

0.05 : 0.05

Mass of K₂SO₄:

Mass = number of moles × molar mass

Mass = 0.05 mol × 174.3 g/mol

Mass = 8.715 g

Thus, from 4.9 g of sulfuric acid 8.175 g of potassium sulfate is formed.

8 0

3 years ago

The area that includes everything between the cell membrane and the nucleus of a cell is called the ___________.