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Elza [17]
3 years ago
12

If 14.5 g of MnO4- (permanganate) react with manganese (II) hydroxide how many grams of manganese (IV) oxide will be produced? T

he other products are hydroxide ion and water.
Chemistry
1 answer:
Veronika [31]3 years ago
5 0

Answer:

m_{MnO_2}=21.2gMnO_2

Explanation:

Hello,

In this case, given the balanced reaction:

2MnO_4^-+2Mn(OH)_2\rightarrow 4MnO_2+2OH^-+H_2O

We can see a 2:4 mole ration between permanganate ion (118.9 g/mol) and manganese (IV) oxide (86.9 g/mol), that is why the resulting mas of this last one turns out:

m_{MnO_2}=14.5gMnO_4^-*\frac{1molMnO_4^-}{118.9gMnO_4^-}*\frac{4MnO_2}{2molMnO_4^-}  *\frac{86.9gMnO_2}{1molMnO_2} \\\\m_{MnO_2}=21.2gMnO_2

Best regards.

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A gas occupies the volume of 215ml at 15C and 86.4kPa?
kherson [118]

Answer:

About 0.1738 liters

Explanation:

Using the formula PV=nRT, where p represents pressure in atmospheres, v represents volume in liters, n represents the number of moles of ideal gas, R represents the ideal gas constant, and T represents the temperature in kelvin, you can solve this problem. But first, you need to convert to the proper units. 215ml=0.215L, 86.4kPa is about 0.8527 atmospheres, and 15C is 288K. Plugging this into the equation, you get:

0.8527\cdot 0.215=n \cdot 0.0821 \cdot 288\\n\approx 7.754 \cdot 10^{-3}

Now that you know the number of moles of gas, you can plug back into the equation with STP conditions:

1V=7.754 \cdot 10^{-3} \cdot 0.0821 \cdot 273\\V\approx 0.1738L

Hope this helps!

3 0
3 years ago
Which equation shows how to calculate how many grams (g) of KOH would be needed to fully react with 4 mol Mg(OH)2? The balanced
bazaltina [42]
Mole ratio:

MgCl₂ + 2 KOH = Mg(OH)₂ + 2 KCl

2 moles KOH ---------------- 1 mole Mg(OH)₂
moles KOH ------------------- 4 moles Mg(OH₂)

moles KOH = 4 x 2 / 1

= 8 moles of  KOH

molar mass KOH = 56 g/mol

mass of KOH = n x mm

mass of  KOH = 8 x 56

= 448 g of KOH 

hope this helps!

3 0
3 years ago
Read 2 more answers
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m
koban [17]

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

R_{in} = \frac{0.04}{100}*2000

R_{in} = 0.8

The rate-out

R_{out} = \frac{A}{6000}*2000

R_{out} = \frac{A}{3}

We can say that:

\frac{dA}{dt}= 0.8-\frac{A}{3}

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

\frac{dA}{dt} +\frac{A}{3} =0.8

Integration of the above linear equation =

e^{\int\limits \frac {1}{3}dt } = e^{\frac{1}{3}t

so we have:

e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A = 0.8e^{\frac{1}{3}t

\frac{d}{dt}[e^{\frac{1}{3}t}A] = 0.8e^{\frac{1}{3}t

Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C

∴ A(t) = 2.4 +Ce^{-\frac{1}{3}t

Since A(0) = 12

Then;

12 =2.4 + Ce^{-\frac{1}{3}}(0)

C= 12-2.4

C =9.6

Hence;

A(t) = 2.4 +9.6e^{-\frac{t}{3}}

A(0) = 2.4 +9.6e^{-\frac{10}{3}}

A(t) = 2.74

∴ the concentration at 10 minutes is ;

=  \frac{2.74}{6000}*100%

= 0.0456667 %

= 0.046% to three decimal places

7 0
3 years ago
What fuel source is Jan using if she exercises at 85% of her maximum aerobic capacity?
Marina86 [1]

Answer:

Carbohydrates

Explanation:

Increased exercise intensity means the overall need for energy increases. As we increase exercise intensity we increase our glucose uptake and oxidation which far exceeds uptake, indicating that muscle stores of glycogen are being used. At moderate intensities (65%) there is an increased need for muscle glycogen and muscle triglycerides which is fat. At higher levels of intensities (85%) there is an even greater need for energy, and this is met almost solely by an increased uptake of glucose from the blood and from muscle glycogen.

In the case of fats as an energy fuel source at high intensities, increasing levels of intensity increases fat oxidation but once we get into higher levels of intensity, we return to levels of fat oxidation similar to very low intensities.

4 0
3 years ago
7. Explain how magnetic potential energy can be transformed into kinetic energy. (1<br> point)
Arisa [49]
The potential energy by the magnetic field can turn into kinetic energy once the field is moving from the S pole to the N pole when it reaches the N pole it is potential energy when it exits the S pole it is kinetic energy.
5 0
1 year ago
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