PLZ HELP ME!!!!!! BEING TIMED!!!!!

According to the reaction represented by the unbalanced equation above, how many moles of so2(g) are required to react completel

melomori [17]

2SO₂ + O₂ = 2SO₃

n(O₂)=1 mol

n(SO₂)=2n(O₂)

n(SO₂)=2 mol

6 0

3 years ago

The breaking buffer that we use this week contains 10mM Tris, pH 8.0, 150mM NaCl. The elution buffer is breaking buffer that als

olasank [31]

Answer:

A breakdown of the breaking buffer was first listed with its respective component and their corresponding value; then a table was made for the stock concentrations in which the volume that is being added was determined by using the formula $M_1*V_1 = M_2*V_2$ . It was the addition of these volumes altogether that make up the 0.25 L (i.e 250 mL) with water

Explanation:

Given data includes:

Tris= 10mM

pH = 8.0

NaCl = 150 mM

Imidazole = 300 mM

In order to make 0.25 L solution buffer ; i.e (250 mL); we have the following component.

Stock Concentration Volume to be Final Concentration

added

1 M Tris 2.5 mL 10 mM

5 M NaCl 7.5 mL 150 mM

1 M Imidazole 75 mL 300 mM

$M_1*V_1 = M_2*V_2$ . is the formula that is used to determine the corresponding volume that is added for each stock concentration

The stock concentration of Tris ( 1 M ) is as follows:

$M_1*V_1 = M_2*V_2$ .

$1*V_1= 0.01 M *250mL\\V_1 = 2.5mL$

The stock concentration of NaCl (5 M ) is as follows:

$M_1*V_1 = M_2*V_2$ .

$1*V_1= 0.15 M *250mL\\V_1 = 7.5mL$

The stock concentration of Imidazole (1 M ) is as follows:

$M_1*V_1 = M_2*V_2$ .

$1*V_1= 0.03 M *250mL\\V_1 = 75mL$

Hence, it is the addition of all the volumes altogether that make up 0.25L (i.e 250 mL) with water.

5 0

3 years ago

If 1.02 g of nickel reacted with 750. mL of 0.112 M hydrobromic acid, how much of each will be present at the end of the reactio

kati45 [8]

Answer:

35.1% is percent yield

Explanation:

Full question: Assume no volume change. If you formed 0.0910 atm of gas, what is the percent yield?

The reaction that is occurring is:

Ni + 3HBr → NiBr₃ + 3/2H₂(g)

First, we will determine moles of Ni and HBr to determine limiting reactant and theoretical yield

Using ideal gas law, we can determine the moles of hydrogen formed. Thus, we can find percent yield:

Moles Ni (Molar mass: 58.69g/mol):

1.02g * (1mol / 58.69g) = 0.01738moles Ni

Moles HBr:

0.750L * (0.112mol/L) = 0.084 moles of HBr.

For a complete reaction of the 0.084 moles of HBr you need:

0.084mol HBr * (1 mole Ni / 3 moles HBr) = 0.028 moles of Ni.

As there are just 0.01738 moles of Ni, the Ni is limiting reactant. Assuming a theoretical yield, moles of H₂ produced are:

0.01738moles Ni * (3/2 H₂ / 1 mol Ni) = 0.02607 moles H₂

Now, moles of H₂ produced are:

PV = nRT

PV/RT = n

Where P is pressure (0.0910atm)

V is volume (2.50L)

R is gas constant (0.082atmL/molK)

T is absolute temperature in Kelvin (30°C + 273.15 = 303.15K)

And n are moles

PV/RT = n

0.0910atm*2.50L/0.082atmL/molK*303.15K = n

0.00915 moles = n

And percent yield (Produced moles / Theoretical moles * 100) is:

0.00915 moles / 0.02607moles =

<h3>35.1% is percent yield</h3>

8 0

3 years ago

adam wants to investigate the strengths of different acids and alkalis. He has 2 different acid solutions and 2 different alkali

SCORPION-xisa [38]

Answer:

The strength of an acid or alkali depends on the degree of dissociation of the acid or alkali in water. The degree of dissociation measures the percentage of acid molecules that ionise when dissolved in water. He could use universal indicators or litmus paper for this.

Explanation:

(See answer for the explanation)

5 0

3 years ago

A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$

calculating the reverse reaction

$H_2(g) + I_2(g)\rightleftharpoons 2HI(g)$

$Kc = \frac{1}{Kc} \\\\$

$= \frac{1}{0.034386}\\ \\= 29.081\\$

7 0

3 years ago