The density of mercury is 13.6g/mL, how much will .5 Liters weigh?

Vilka [71]

As iron heats up, the arrangement of its atoms changes several times before it melts. ... Iron is magnetic at room temperature, and previous work predicted that iron's magnetism favors its open structure at low temperatures, but at 770 degrees Celsius iron loses its magnetism.

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8 0

3 years ago

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How many molecules of carbon dioxide are in 9.080 x 10 ^-1 mol?

kakasveta [241]

Answer:

5.46 8 x 10²³ molecules.

Explanation:

Knowing that every one mole of a substance contains Avogadro's no. of molecules (NA = 6.022 x 10²³).

Using cross multiplication:

1.0 mole → 6.022 x 10²³ molecules.

9.08 x 10⁻¹ mole → ??? molecules.

∴ The no. of molecules of CO₂ are in 9.08 x 10⁻¹ mol = (6.022 x 10²³ molecules) ( 9.08 x 10⁻¹ mole) / (1.0 mol) = 5.46 8 x 10²³ molecules.

6 0

3 years ago

At 2°C, the vapor pressure of pure water is 23.76 mmHg and that of a certain seawater sample is 23.09 mmHg. Assuming that seawat

wariber [46]

Answer:

0.808 M

Explanation:

Using Raoult's Law

$\frac{P_s}{Pi}= x_i$

where:

$P_s$ = vapor pressure of sea water( solution) = 23.09 mmHg

$P_i$ = vapor pressure of pure water (solute) = 23.76 mmHg

$x_i$ = mole fraction of water

∴

$\frac{23.09}{23.76}= x_i$

$x_i = 0.9718$

$x_i+x_2=1$

$x_2 = 1- x_i$

$x_2 = 1- 0.9718$

$x_2 = 0.0282$

$x_i = \frac{n_i}{n_i+n_2}$ ------ equation (1)

$x_2 = \frac{n_2}{n_i+n_2}$ ------ equation (2)

where; $(n_2) =$ number of moles of sea water

$(n_i) =$ number of moles of pure water

equating above equation 1 and 2; we have :

$\frac{n_2}{n_i}$ $= \frac{0.0282}{0.9178}$

$= 0.02901$

NOW, Molarity = $\frac{moles of sea water}{mass of pure water }*1000$

$= \frac{0.02901}{18}*1000$

$= 0.001616*1000$

$= 1.616 M$

As we assume that the sea water contains only NaCl, if NaCl dissociates to Na⁺ and Cl⁻; we have $\frac{1.616}{2} =0.808 M$

3 0

2 years ago

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If a “sample” of pennies contained 75 heads and 25 tails, how many half‐lives would have passed since the “sample” formed. Expla

finlep [7]

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8 0

2 years ago

What is the molariity of a 50.0 mL aqueous solution containing 10.0 grams of hydrogen peroxide, H2O2?

daser333 [38]

Answer:

6 mol/L

Explanation:

You should know or have the equation to solve for Molarity which is;

M = n/v (M: Molarity) (n: moles of solute) (v: Liters of solute)

You can start off differently but I would start by converting the mL to L. This is your "v" value.

50.0 mL/ 1000 mL = 0.05 L

Now, you have to convert grams to moles in order to solve for molarity (M).

1.) On the periodic table find the molecular weights of H and O.

H= 1.01 g/mol O= 16.00 g/mol

2.) Multiply them and then add them together to have their combined molecular weights. (You have to multiply by 2 because of their equation; H2O2).

2(1.01) + 2(16.00)= 34.02 g/mol

3.) Now, you're going to use the "picket fence method" or whichever your teacher taught you to convert from grams to moles. This will be your "n" value. (I cannot show it on here without it looking weird, so my sincere apologies.)

10.0 g/ 34.02 g = 0.2939 mol

4.)You are now going to plug in your answers into the equation for Molarity.

M= 0.2939 mol / 0.05 L = 5.878 mol/L

5.) I am sure your professor might be a stickler so for sig figs sake when you multiply or divide use the smallest amount of sig figs you see which is 1. Round 5.878 to 6 mol/L

Sorry this explanation is very long let me know if you need a better more written out explanation.

4 0

3 years ago