Consider the incomplete reaction below.

What is the volume of 60 g of ether if the density of ether is 0.70g/mL

miss Akunina [59]

Answer:

114mL.

Explanation: hope this helped

8 0

3 years ago

What is the oxidation state of S in so ??

Marina86 [1]

Answer:

+2

Explanation:

If a compound $SO$ existed, we would identify the oxidation state of sulfur using the following logic:

oxygen is more electronegative than sulfur, so it's more electron-withdrawing and it should have a negative oxidation state producing a positive oxidation state for sulfur;
oxygen typically has an oxidation state of -2;
we may then apply the fact that SO is expected to be a molecule with a net charge of 0;
if the net charge is 0 and the oxidation state of oxygen is -2, we may set the oxidation state of S to x;
write the equation for the net charge of 0 by adding all individual charges of the two atoms: $x + (-2) = 0$ ;
hence, x = 2.

That said, in this hypothetical compound S would have an oxidation state of +2.

6 0

3 years ago

2. A sample of gas has a pressure of 1.4 atm and a volume of 500 mL. If the volume is decreased to 250 mL, what is the new press

aivan3 [116]

Answer: The new pressure is 2.8 atm.

Explanation:

Since our given information only involves pressure and volume, we will use Boyle's Law: $P_{1} V_{1} =P_{2} V_{2}$ . Let's input the data into the formula.

$1.4atm*500mL=P_{2} *250mL$

We have to rearrange our data a bit now because the volume is supposed to be in liters. To convert from milliliters(mL) to liters(L), you must divide the number by 1000. That is what we will do with both of our given volumes.

$\frac{500mL}{1000} =0.5L$ $\frac{250mL}{1000} =0.25L$

Now that our volumes are in liters, we can correctly plug in our data.

$1.4atm*0.5L=P_{2} *0.25L$

Since we do not know the value $P_{2}$ , we must rearrange the formula into a stoichiometric equation to find out the answer.

$P_{2}=\frac{1.4atm*0.5L}{0.25L}$ $P_{2}=2.8 atm$

Our final answer is 2.8 atmospheres.

I hope this helps!! Pls mark brainliest :))

5 0

2 years ago

Aqueous potassium iodate and potassium iodide react in the presence of dilute hydrochloric acid, as shown below. kio3(aq) + 5ki(

Musya8 [376]

<span>0.0797 g Looking at the formula, 1 mole of KIO3 and 5 moles of KI will react and produce moles of iodine molecules or 6 moles of iodine atoms. So first, determine the number of moles of KIO3 and KI provided moles KIO3 = 0.0121 * 0.097 = 0.0011737 mol moles KI = 0.0308 * 0.017 = 0.0005236 mol The limiting reactant is KI at 0.0005236 mol so divide by 5 and multiply by 6 to get the number of moles of iodine atoms. 0.0005236 / 5 * 6 = 0.00062832 mol Lookup the atomic weight of iodine which is 126.90447 And multiply that by the number of moles of iodine produced 126.90447 g/mol * 0.00062832 mol = 0.079736617 g Rounding to 4 decimal places gives 0.0797 g</span>

4 0

3 years ago

If atmospheric pressure on a certain day is 749 mmHg, what is the partial pressure of nitrogen, given that nitrogen is about 78%

Gekata [30.6K]

Answer: 584 mm Hg

Explanation:

According to Raoult's law, the partial pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the total pressure.

$p_1=x_1P$

where, x = mole fraction

$P$ = total pressure = 749 mmHg

$x_{N_2}=\frac{\text {moles of }N_2}{\text {total moles}}=\frac{78}{100}=0.78$

Putting in the values we get:

$p_1=0.78\times 749mmHg=584mmHg$

The partial pressure of nitrogen will be 584 mmHg

4 0

3 years ago