How many mililiters is 0.55 L

A solution with pH of 9 has [OH-] concentration of

Drupady [299]

PH = -log([H+])
[H+] = 10^(-pH)

[H+] = 10^(-9)

[H+][OH-] = Kw
Kw = 1.0*10^-14 at 25 degrees celsius.

[OH-] = Kw/[H+] = (1.0*10^-14)/(1*10^-9) = 1.0*10^-5

The concentration of OH- ions is 1.0*10^-5 M.

7 0

3 years ago

Which organism has a nerve net? A fish B hydra C flatworm D sponge

Kaylis [27]

B. Hydra, because Hydra<span>are an ideal class of </span>Cnidaria<span> to research and on which to run tests.</span>

5 0

3 years ago

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The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process.... In the first step, manganes

frez [133]

Answer : The mass of $MnCO_3$ required are, 35 kg

Explanation :

First we have to calculate the mass of $MnO_2$ .

The first step balanced chemical reaction is:

$2MnCO_3+O_2\rightarrow 2MnO_2+2CO_2$

Molar mass of $MnCO_3$ = 115 g/mole

Molar mass of $MnO_2$ = 87 g/mole

Let the mass of $MnCO_3$ be, 'x' grams.

From the balanced reaction, we conclude that

As, $(2\times 115)g$ of $MnCO_3$ react to give $(2\times 87)g$ of $MnO_2$

So, $xg$ of $MnCO_3$ react to give $\frac{(2\times 87)g}{(2\times 115)g}\times x=0.757xg$ of $MnO_2$

And as we are given that the yield produced from the first step is, 65 % that means,

$60\% \text{ of }0.757xg=\frac{60}{100}\times 0.757x=0.4542xg$

The mass of $MnO_2$ obtained = 0.4542x g

Now we have to calculate the mass of $Mn$ .

The second step balanced chemical reaction is:

$3MnO_2+4Al\rightarrow 3Mn+2Al_2O_3$

Molar mass of $MnO_2$ = 87 g/mole

Molar mass of $Mn$ = 55 g/mole

From the balanced reaction, we conclude that

As, $(3\times 87)g$ of $MnO_2$ react to give $(3\times 55)g$ of $Mn$

So, $0.4542xg$ of $MnO_2$ react to give $\frac{(3\times 55)g}{(3\times 87)g}\times 0.4542x=0.287xg$ of $Mn$

And as we are given that the yield produced from the second step is, 80 % that means,

$80\% \text{ of }0.287xg=\frac{80}{100}\times 0.287x=0.2296xg$

The mass of $Mn$ obtained = 0.2296x g

The given mass of Mn = 8.0 kg = 8000 g (1 kg = 1000 g)

So, 0.2296x = 8000

x = 34843.20 g = 34.84 kg = 35 kg

Therefore, the mass of $MnCO_3$ required are, 35 kg

4 0

3 years ago

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What volume would 56.2 mL of gas at 820 mm of Hg occupy at 720 mm of Hg?

Andreyy89

Answer:

49.35 mL

Explanation:

Given: 56.2 mL of gas

To find: volume that 56.2 mL of gas at 820 mm of Hg would occupy at 720 mm of Hg

Solution:

At 820 mm of Hg, volume of gas is 56.2 mL

At 1 mm of Hg, volume of gas is $\frac{56.2}{820}$

At 720 mm of Hg, volume of gas is $\frac{56.2}{820}(720)=49.35\,\,mL$

3 0

3 years ago

In one to two sentences, describe an experiment that would show that intramolecular forces (attractions between atoms within mol

Tatiana [17]

An experiment that would show that intramolecular forces are stronger than intermolecular forces will be heating a block of ice in a sealed container then allowing it to change to steam.

Intramolecular forces are the forces of attraction that hold atoms together within a molecule. Intramolecular forces require a high amount of energy to splits atoms or molecules in a chemical bonding.

Intermolecular forces are weaker forces of attraction that occur between molecules. They require lesser energy to splits molecules compared to intramolecular forces.

An experiment that would show that intramolecular forces are stronger than intermolecular forces will be heating a block of ice in a sealed container then allowing it to change to steam.

In the process, the energy required to change the state from ice to steam water is more than intermolecular forces.

Thus, we can conclude that this experiment shows that the intramolecular forces are stronger than the intermolecular forces.

Learn more about Intramolecular forces here:

brainly.com/question/13588164

3 0

2 years ago