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3 years ago

How many grams are there in 1.70 moles of KMnO4?

Chemistry

2 answers:

topjm [15]3 years ago

8 0

Answer:

269g

Explanation:

hope this helps you

Rashid [163]3 years ago

6 0

Answer: 269 g
Explanation: since we want moles>grams you would multiply the given by the molar mass of the element divided by one.So 1.70*158/1=269 grams

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When the following equation is balanced with lowest whole-number coefficients, what is the coefficient for NO(g)? ___NH3(g) + __

inn [45]

Answer:

Coefficient in front of the $NO$ in the balanced equation - 4

Explanation:

The unbalanced reaction is shown below as:-

$NH_3+O_2\rightarrow NO+H_2O$

On the left hand side,

There are 1 nitrogen atom and 3 hydrogen atoms and 3 oxygen atoms

On the right hand side,

There are 1 nitrogen atom and 2 hydrogen atoms and 2 oxygen atoms

Thus,

Right side, $H_2O$ must be multiplied by 6 and $NO$ by 4

Left side, $O_2$ is multiplied by 5 and $NH_3$ by 4 so to balance the whole reaction.

Thus, the balanced reaction is:-

$4NH_3+5O_2\rightarrow 4NO+6H_2O$

<u>Coefficient in front of the $NO$ in the balanced equation - 4</u>

5 0

3 years ago

Question 1 A student made measurements on some electrochemical cells and calculated three quantities: The standard reaction free

Zinaida [17]

Answer:

Explanation:

for spontaneous reaction,

ΔG is negative

K>1

E > 0

cell A:

ΔG and EO suggests that reaction is spontaneous. But K is less than 1.

Hence K is wrong

cell B:

ΔG and EO suggests that reaction is non spontaneous .But K is greater than 1.

Hence K is wrong

cell C:

E and K suggest than reaction is non spontaneous but ΔG suggest that reaction is spontaneous.

Hence ΔG is wrong

4 0

4 years ago

Read 2 more answers

What is the pH of 4.3x10^-7 M solution of H2CO3?

Rzqust [24]

Answer:

pH = -log[H⁺] = -log(4.75 x 10⁻⁷) = -(- 6.32 ) = 6.32

Explanation:

Given 4.3 x 10⁻⁷M H₂CO₃ (Ka1 = 4.2 x 10⁻⁷ & Ka2 = 4.8 x 10⁻¹¹)

Note: The Ka2 value for the 2nd ionization step is so small (Ka2 = 4.8 x 10⁻¹¹) It will be assumed all of the hydronium ions (H⁺) come from the 1st ionization step.

1st Ionization step

H₂CO₃ ⇄ H⁺ + HCO₃⁻

C(initial) 4.3 x 10⁻⁷ 0 0

ΔC -x +x +x

C(final) 4.3 x 10⁻⁷ - x x x

Note: the 'x' value in this analysis can not be dropped as the Conc/Ka value is less than 10². In this case he C/Ka ratio* (4.3E-7/4.2E-7 ≈ 1) is far below 10².

So, one sets up the equilibrium equation to be quadric and the x-value can be determined.

Ka1 = [H⁺][HSO⁻]/[H₂SO₃] = (x)(x)/(4.3 x 10⁻ - x) = x²/(4.3 x 10⁻⁷ - x) = 4.2 x 10⁻⁷

=> x² = 4.2 x 10⁻⁷(4.3 x 10⁻⁷ - x)

=> x² + 4.2 x 10⁻⁷x - 1.8 x 10⁻¹³ = 0

a = 1, b = 4.2 x 10⁻⁷, c = - 1.8 x 10⁻¹³

x = b² ± SqrRt(b² - 4(1)(-1.8 x 10⁻¹³ / 2(1) = 4.75 x 10⁻⁷

x = [H⁺] = 4.75 x 10⁻⁷

pH = -log[H⁺] = -log(4.75 x 10⁻⁷) = -(- 6.4 ) = 6.4

______________________

* The Concentration/Ka-value is the simplification test for quadratic equations used in Equilibrium studies. If the C/Ka > 100 then one can simplify the C(final) by dropping the 'x' if used in this type analysis. However, if the C/Ka value is < 100 then the x-value must be retained and the solution is determined using the quadratic equation formula.

for ax² + bx + c = 0

x = b² ± SqrRt(b² - 4ac) / 2a

4 0

3 years ago

How many liters of chlorine gas at 25°C and 0.950 atm can be produced by the reaction of 12.0 g of MnO2 with excess HCl(aq) acco

iragen [17]

Answer:

3.55 L.

Explanation:

We'll begin by calculating the number of mole in 12 g of MnO2. This can be obtained as follow:

Molar mass of MnO2 = 55 + (16×2)

= 55 + 32

= 87 g/mol

Mass of MnO2 = 12 g

Mole of MnO2 =...?

Mole = mass /Molar mass

Mole of MnO2 = 12 / 87

Mole of MnO2 = 0.138 mole

Next, we shall determine the number of mole Cl2 produced from the reaction. This is illustrated below:

The balanced equation for the reaction is given below:

MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l) + Cl2(g)

From the balanced equation above,

1 mole of MnO2 reacted to produce 1 mole of Cl2.

Therefore, 0.138 mole of MnO2 will also produce 0.138 mole of Cl2.

Finally, we shall determine the volume of Cl2 gas obtained from the reaction. This can be obtained as shown below:

Temperature (T) = 25 °C = 25 °C + 273 = 298 K

Pressure (P) = 0.950 atm

Number of mole (n) = 0.138 mole

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) =.?

PV = nRT

0.950 × V = 0.138 × 0.0821 × 298

Divide both side by 0.950

V = (0.138 × 0.0821 × 298) / 0.950

V = 3.55 L

Therefore, 3.55 L of chlorine gas were obtained from reaction.

4 0

3 years ago

What coefficients would balance the following equation?<br><br> __C2H6 + __O2 → __CO2 + __H2O

lina2011 [118]

2C2H6 + 7O2 → 4CO2 + 6H2O

3 0

3 years ago