All categories

2 years ago

Help with chemistry, 50 points

Chemistry

2 answers:

expeople1 [14]2 years ago

4 0

Answer:

Thank you

I didn't wait because there is someone who will take it and there is someone who answer a wrong answer

Explanation:

Aloiza [94]2 years ago

3 0

Answer:

Poopoo

Explanation:

You might be interested in

What ion is at a higher concentration in the extracellular fluid than in the intracellular fluid?

Phoenix [80]

There are two types of fluid in the body extracellular fluid and intracellular fluid (ECF and ICF), together they are account for total body water.
The Sodium (Na+) ion is at higher concentration in the extracellular fluid than in the intracellular fluid. The function of extracellular fluid is that it provide cells to watery environment so that they can easily live and perform their function.

8 0

2 years ago

Cci(4) will be soluble in

Romashka-Z-Leto [24]

LIKE DISSOLVES LIKE. Since Ccl4 is non-polar, it'll be soluble in any non-polar solvent. Hope this helps you!

8 0

3 years ago

Calculate the vapor pressure (in torr) of a 35°C solution of ethanol and propanol where the mole fraction of propanol is .75.

astraxan [27]

Answer:

i wish i know

Explanation:

4 0

2 years ago

PLS help

viktelen [127]

Answer:

d . Sc2O5

Explanation:

Hello,

In this case, when forming oxides from a metal and oxygen, for us to find out each element's subscript, we must exchange them as shown below, considering +5 for scandium:

$Sc^{+5}_2O^{-2}_5$

For that reason, the answer is d . Sc2O5

Best regards.

3 0

3 years ago

Read 2 more answers

How many moles of ScCl3 can be produced when 10.00 mol Sc react with 9.00 mol Cl2

Nuetrik [128]

Answer:

Moles of $ScCl_3$ = 6 moles

Explanation:

The reaction of $Sc$ and $Cl_2$ to make $ScCl_3$ is:

$2Sc+3Cl_2$ ⇒ $2ScCl_3$

The above reaction shows that 2 moles of Sc can react with 3 moles of $Cl_2$ to form $ScCl_3.$

Mole Ratio= 2:3

For 10 moles of Sc we need:

Moles of $Cl_2$ = $Moles of Sc *\frac{3 moles of Cl_2}{2 Moles of Sc}$

Moles of $Cl_2$ = $10 *\frac{3 moles of Cl_2}{2 Moles of Sc}$

Moles of $Cl_2$ =15 moles

So 15 moles of $Cl_2$ are required to react with 10 moles of $Sc$ but we have 9 moles of $Cl_2$ , it means $Cl_2$ is limiting reactant.

$Moles of ScCl_3=Given\ Moles\ of\ Cl_2 *\frac{2\ Moles\ o\ fScCl_3}{3\ Moles\ of\ Cl_2}$

$Moles\ of\ ScCl_3=9 *\frac{2\ Moles\ of\ ScCl_3}{3\ Moles\ of\ Cl_2}$

Moles of ScCl_3= 6 moles

4 0

2 years ago

Read 2 more answers