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Lesechka [4]
3 years ago
6

What is the density of an object with the density of 3 when cut in half?

Chemistry
1 answer:
anyanavicka [17]3 years ago
8 0
The density is 3 because the density remains the same.
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1‑propanol ( n ‑propanol) and 2‑propanol (isopropanol) form ideal solutions in all proportions. Calculate the partial pressure a
Alex777 [14]

Answer:

y_{prop} = 0.134; y_{iso} = 0.866

The partial pressure of isopropanol = 34.04 Torr; The partial pressure of propanol = 5.26 Torr

Explanation:

For each of the solutions:

mole fraction of isopropanol  (x_{iso}) = 1 - mole fraction of propanol (x_{prop}).

Given: mole fraction of propanol = 0.247. Thus, the mole fraction of isopropanol = 1 - 0.247 = 0.753.

Furthermole, the partial pressure of isopropanol = x_{iso}*vapor pressure of isopropanol = 0.753*45.2 Torr = 34.04 Torr

The partial pressure of propanol = x_{prop}*vapor pressure of propanol = 0.247*20.9 Torr = 5.16 Torr

Similarly,

In the vapor phase,

The mole fraction of propanol (y_{prop}) = \frac{P_{prop} }{P_{prop}+P_{iso}}

Where, P_{prop} is the partial pressure of propanol and P_{iso} is the partial pressure of isopropanol.

Therefore,

y_{prop} = 5.26/(34.04+5.16) = 0.134

y_{iso} = 1 - 0.134 = 0.866

5 0
3 years ago
Which of the following equations are correctly balanced?
AlladinOne [14]

i think it's I

I was confused by IV then search on gg and it said ZnSO4 should be Zn2SO4 instead but still im not sure Zn2SO4 is real

3 0
3 years ago
What do protons determine about an element
Daniel [21]
The number of protons in an atoms determines the atoms identity. Electrons determine the electrical charge.
5 0
3 years ago
Read 2 more answers
Which element decreases its oxidation number in this reaction? bicl2 + na2so4 → 2nacl + biso4?
kifflom [539]
The balanced reaction is as follows;
BiCl₂ + Na₂SO₄ --> 2NaCl + BiSO₄
this is a double displacement reaction 
the oxidation number of Bi is +2 in both BiCl₂ and BiSO₄
oxidation number of Cl is -1 in both BiCl₂ and NaCl 
oxidation number of Na is +1 in both Na₂SO₄ and NaCl
oxidation numbers of elements in SO₄²⁻ remains the same in both compounds.Therefore the oxidation state in any of the elements in the reaction doesn't change. Neither of the elements show an increase or decrease in the oxidation numbers .
Answer for this question is no element decreases its oxidation number.

5 0
3 years ago
Read 2 more answers
At a certain temperature the vapor pressure of pure chloroform (CHCl3) is measured to be 91. torr. Suppose a solution is prepare
OleMash [197]

Answer:

P_{CCl_4}=52.43torr

Explanation:

Hello there!

In this case, sine the solution of this problem require the application of the Raoult's law, assuming heptane is a nonvolatile solute, so we can write:

P_{CCl_4}=x_{CCl_4}P_{CCl_4}^{vap}

Thus, we first calculate the mole fraction of chloroform, by using the given masses and molar masses as shown below:

x_{CCl_4}=\frac{140/153.81}{140/153.81+67.1/100.21}=0.576

Therefore, the partial pressure of chloroform turns out to be:

P_{CCl_4}=0.576*91torr\\\\P_{CCl_4}=52.43torr

Regards!

3 0
3 years ago
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