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kogti [31]
2 years ago
8

The label on a container of margarine lists "hydrogenated vegetable oil" as the major ingredient. Hydrogenated vegetable oil ___

_____.
Chemistry
1 answer:
irina1246 [14]2 years ago
5 0

Answer:

The hydrogenated vegetable oil stays solid at room temperature

Explanation:

According to my research on studies conducted by various chemists, I can say that based on the information provided within the question the hydrogenated vegetable oil allows the margarine to be able to stay solid at room temperature. Otherwise the margarine would melt at room temperature and virtually become unusable or unsafe to consume.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

You might be interested in
2.1 liters is the same as cm3 and ml
Kipish [7]
1L = 1000ml
1ml = 1cm^3

2.1L = 2100mL = 2100cm^3

4 0
2 years ago
If an aqueous solution of urea n2h4co is 26% by mass and has density of 1.07 g/ml, calculate the molality of urea in the soln
Ksju [112]
Answer is: molality of urea is 5.84 m.

If we use 100 mL of solution:
d(solution) = 1.07 g/mL.
m(solution) = 1.07 g/mL · 100 mL.
m(solution) = 107 g.
ω(N₂H₄CO) = 26% ÷ 100% = 0.26.
m(N₂H₄CO) = m(solution) · ω(N₂H₄CO).
m(N₂H₄CO) = 107 g · 0.26.
m(N₂H₄CO) = 27.82 g.
1) calculate amount of urea:
n(N₂H₄CO) = m(N₂H₄CO) ÷ M(N₂H₄CO).
n(N₂H₄CO) = 27.82 g ÷ 60.06 g/mol.
n(N₂H₄CO) = 0.463 mol; amount of substance.
2) calculate mass of water:
m(H₂O) = 107 g - 27.82 g.
m(H₂O) = 79.18 g ÷ 1000 g/kg.
m(H₂O) = 0.07918 kg.
3) calculate molality:
b = n(N₂H₄CO) ÷ m(H₂O).
b = 0.463 mol ÷ 0.07918 kg.
b = 5.84 mol/kg.
5 0
2 years ago
Read 2 more answers
What is the fraction of the hydrogen atom's mass (11h) that is in the nucleus? the mass of proton is 1.007276 u, and the mass of
ololo11 [35]
Hello there!

To determine the fraction of the hydrogen atom's mass that is in the nucleus, we have to keep in mind that a Hydrogen atom has 1 proton and 1 electron. Protons are in the nucleus while electrons are in electron shells surrounding the nucleus. The mass of the nucleus will be equal to the mass of 1 proton and we can express the fraction as follows:

Mass Fraction= \frac{mass 1 Proton}{mass H atom} = \frac{1,007276 u}{1,007825}=0,9995

So, the fraction of the hydrogen atom's mass that is in the nucleus is 0,9995. That means that almost all the mass of this atom is at the nucleus.

Have a nice day!
3 0
2 years ago
In Universe L, recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, quan
Advocard [28]

Answer:

Manganese, Fifth transition element

[X] 3d⁶ 4s¹

Iron, Sixth transition element

[X] 3d⁶ 4s²

Explanation:

Complete Question

In Universe L, recently discovered by an intrepid team of chemists who also happen to have studied interdimensional travel, quantum mechanics works as it does in our universe, except that there are six d orbitals instead of the usual number we observe here. Use these facts to write the ground-state electron configurations of the sixth and seventh elements in the first transition series in Universe L. Note; you may use [X] to stand for the electron configuration of the noble gas at the end of the row before the first transition series.

Solution

In our universe, there are 5 d orbitals.

And according to Aufbau's principles that electrons fill the lower energy orbitals before they fill higher energy orbitals and Hund's Rule that states that electrons are fed singly to all the orbitals of a subshell before pairing occurs.

The fifth and sixth transition elements in our universe is then Manganese and Iron respectively.

Manganese - [Ar] 3d⁵ 4s²

Iron - [Ar] 3d⁶ 4s²

So, in the new universe L, where there are six d orbitals, for manganese, the fifth transition metal, because half filled orbitals are more stable than partially filled orbitals (that woukd have been rhe case if we leave 5 electrons on the 3d orbital), the 4s orbital is filled to half of its capacity and the one electron removed from the 4s is used to fill the six 3d orbital to half of its capacity too.

For the sixth transition element, the new extra electron just fills the lower energy 4s orbital, leaving the six 3d orbitals all half-filled.

Hence, they both have ground state configurations of

- Manganese, Fifth transition element

[X] 3d⁶ 4s¹

- Iron, Sixth transition element

[X] 3d⁶ 4s²

Hope this Helps!!!

7 0
2 years ago
A student heats a sample of Copper (II) sulfate in a crucible and records the data shown in the table. What is the complete form
liberstina [14]

Explanation:

Copper (II) sulfate is usually present as a hydrous state, which is of the form CuSO4 * nH2O, where n is a whole number.

Mass of sample (CuSO4 * nH2O)

= 152.00g - 128.10g = 23.90g.

Mass of water loss during heating

= 152.00g - 147.60g = 4.40g.

Molar mass of H2O = 18g/mol

Moles of H2O in sample

= 4.40g / (18g/mol) = 0.244mol.

Mass of anhydrous sample (CuSO4)

= 23.90g - 4.40g = 19.50g

Molar mass of CuSO4 = 159.61g/mol

Moles of CuSO4 in sample

= 19.50g / (159.61g/mol) = 0.122mol.

Since mole ratio of CuSO4 to H2O

= 0.122mol : 0.244mol = 1:2, n = 2.

Hence we have CuSO4 * 2H2O.

6 0
2 years ago
Read 2 more answers
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