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2 years ago

13

A gas is collected at a temperature of 25.2 degrees celcius and a pressure of 0.967 atm. What pressure would the gas have at a t

emperature of 44.5 degrees celcius?

1 answer:

Nimfa-mama [501]2 years ago

5 0

Answer:

1.707599191 atm

Explanation:

44.5 divided by 25.2 = 1.765873 (i used calculator to double check)

1.765873 x 0.967 = 1.707599191 (used calculator)

You might be interested in

copper hydroxide and potassium sulfate are produced when potassium hydroxide reacts with copper sulfate balanced equation

a_sh-v [17]

This problem is requiring the balanced chemical equation that takes place when copper hydroxide and potassium sulfate are produced when reacting potassium hydroxide with copper sulfate.

<h3>Balancing chemical equations:</h3>

In chemistry, balancing chemical equations is based on the law of conservation of mass, which demands us to have equal number of atoms on both sides of the chemical equation. This can be accomplished by inserting coefficients in front of the chemical species.

For this particular case, we have potassium hydroxide with copper sulfate on the reactants side, however, copper can be copper (I) or copper (II) as it has 1+ and 2+ as its possible oxidation numbers. In addition, copper hydroxide and potassium sulfate as the products. Hence, we can assume this is all about copper (II) so we can write:

$KOH+CuSO_4\rightarrow K_2SO_4+Cu(OH)_2$

As we can see, potassium, hydrogen and oxygen have two atoms each on the products side, but just one on the reactants side; drawback we can overcome by putting a 2 in front of KOH so as to balance it:

$2KOH+CuSO_4\rightarrow K_2SO_4+Cu(OH)_2$

Learn more about balancing chemical equations: brainly.com/question/8062886

8 0

2 years ago

In the laboratory you dissolve 22.0 g of nickel(II) nitrate in a volumetric flask and add water to a total volume of 125

Ahat [919]

Answer:

<u>1.364 M</u>.

Explanation:

Molarity formula: M= n/v, where n is moles of solute, and v is liters of solution.

Now, we need to convert the grams of nickel to moles and the volume of water to liters.

125mL/1000= 0.125 L.

To convert nickel grams to moles, we need to take a look at it's chemical formula, which is:

$Ni(NO_{3} )_{2}$

Now we count how many molecules of each element we have:

Ni= 1

N= 2

O= 6

Calculate the weight (g) of each element (the values of g/mol can be found on the periodic table and they may vary slightly between one table and the other):

Ni: (1) (58.6934)= 58.6934

N= (2) (14.007)= 28.014

O= (6) (15.999)= 95.994

Sum all the values to obtain the total weight of 1 mole of this compound:

58.6934+28.014+95.994= 129(g/mole)

Now that we know the that 129 grams equal 1 mole of nickel(II) nitrate, we can convert the 22.0 g to moles:

129g ------- 1 mole

22.0g ----- x

x= (22*1)/129= 0.1705 moles.

Now, we have all the values needed to calculate the molarity of this solution. All we have to do is substitute the values in the formula:

M= (0.1705 moles) / (0.125 L)= <u>1.364 M</u>.

3 0

2 years ago

chemistry A basketball is inflated to a pressure of 1.10 atm in a 28.0°C garage. What is the pressure of the basketball outside

butalik [34]

Answer : The final pressure of the basketball is, 0.990 atm

Explanation :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

$P\propto T$

or,

$\frac{P_1}{P_2}=\frac{T_1}{T_2}$

where,

$P_1$ = initial pressure = 1.10 atm

$P_2$ = final pressure = ?

$T_1$ = initial temperature = $28.0^oC=273+28.0=301.0K$

$T_2$ = initial temperature = $-2.00^oC=273+(-2.00)=271.0K$

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

$\frac{1.10atm}{P_2}=\frac{301.0K}{271.0K}$

$P_2=0.990atm$

Thus, the final pressure of the basketball is, 0.990 atm

7 0

3 years ago

Picric acid has been used in the leather industry and in etching copper. However, its laboratory use has been restricted because

olchik [2.2K]

Answer:

0.3023 M

Explanation:

Let Picric acid = $H_{picric}$

So, $H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

The ICE table can be given as:

$H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

Initial: 0.52 0 0

Change: - x + x + x

Equilibrium: 0.52 - x + x + x

Given that;

acid dissociation constant ( $K_a$ ) = 0.42

$K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}$

$0.42 = \frac{[x][x]}{0.52-x}}$

$0.42 = \frac{[x]^2}{0.52-x}}$

0.42(0.52-x) = x²

0.2184 - 0.42x = x²

x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)

Using the quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$ ; ( where +/- represent ± )

= $\frac{-0.42+/-\sqrt{(0.42)^2-4(1)(-0.2184)} }{2*1}$

= $\frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}$

= $\frac{-0.42+\sqrt {1.0496} }{2}$ OR $\frac{-0.42-\sqrt {1.0496} }{2}$

= $\frac{-0.42+1.0245}{2}$ OR $\frac{-0.42-1.0245}{2}$

= $\frac{0.6045}{2}$ OR $-\frac{1.4445}{2}$

= 0.30225 OR - 0.72225

So, we go by the +ve integer that says:

x = 0.30225

x = [ $H_3}O^+$ ] = [ $Picric^-$ ] = 0.3023 M

∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).

6 0

3 years ago

CH3 + HCl <=> CH3Cl + H2O

dmitriy555 [2]

Answer:

The pressure of CH3OH and HCl will decrease.

The final partial pressure of HCl is 0.350038 atm

Explanation:

Step 1: Data given

Kp = 4.7 x 10^3 at 400K

Pressure of CH3OH = 0.250 atm

Pressure of HCl = 0.600 atm

Volume = 10.00 L

Step 2: The balanced equation

CH3OH(g) + HCl(g) <=> CH3Cl(g) + H2O(g)

Step 3: The initial pressure

p(CH3OH) = 0.250atm

p(HCl) = 0.600 atm

p(CH3Cl)= 0 atm

p(H2O) = 0 atm

Step 3: Calculate the pressure at the equilibrium

p(CH3OH) = 0.250 - X atm

p(HCl) = 0.600 - X atm

p(CH3Cl)= X atm

p(H2O) = X atm

Step 4: Calculate Kp

Kp = (pHO * pCH3Cl) / (pCH3* pHCl)

4.7 * 10³ = X² /(0.250-X)(0.600-X)

X = 0.249962

p(CH3OH) = 0.250 - 0.249962 = 0.000038 atm

p(HCl) = 0.600 - 0.249962 = 0.350038 atm

p(CH3Cl)= 0.249962 atm

p(H2O) = 0.249962 atm

Kp = (0.249962 * 0.249962) / (0.000038 * 0.350038)

Kp = 4.7 *10³

The pressure of CH3OH and HCl will decrease.

The final partial pressure of HCl is 0.350038 atm

4 0

2 years ago