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3 years ago

Срочноооооооооооооо!!!!

Chemistry

1 answer:

Ymorist [56]3 years ago

4 0

Answer:

I can't answer this as I am not able to read the Question

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Five gases combined in a gas cylinder have the following partial pressures: 3.00 atm (N2), 1.80 atm (O2), 0.29 atm (Ar), 0.18 at

levacccp [35]

In order to solve the total pressure that is exerted by the gases, we need to use the Dalton's Law of Partial pressures. These are the calculations that you need to find out the total amount of pressure exerted to the gases:

3.00atm (N2) + 1.80atm (O2) + 0.29atm (Ar) + 0.18atm (He) + 0.10atm (H),
add up all of that, and the answer would turn out to be: 5.37atm.

8 0

3 years ago

Read 2 more answers

When butter is heated it melts and when that melted butter cools and solidifies the process called

avanturin [10]

<span>I'm pretty sure it is called condensation</span>

3 0

3 years ago

What mass of radon is in 8.17 moles of radon?

Len [333]

Answer:

1813.74g

Explanation:

Given parameters:

Number of moles of radon = 8.17moles

Unknown:

Mass of radon = ?

Solution:

To solve this problem, we use the expression below:

Number of moles = $\frac{mass}{molar mass}$

Molar mass of radon = 222g/mol

Now insert the parameters and solve;

Mass of radon = Number of moles x molar mass

= 8.17 x 222

= 1813.74g

4 0

3 years ago

Four animals were fleeing from a wildfire. All of them were running at a speed of 30 miles per hour. Which one has the greatest

RUDIKE [14]

Answer: I don’t know lol

Explanation: I am so sorry I thought this was easy

6 0

3 years ago

Read 2 more answers

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

Veronika [31]

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.

5 0

3 years ago