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Aneli [31]
3 years ago
11

A chemist adds of a copper(II) sulfate solution to a reaction flask. Calculate the mass in kilograms of copper(II) sulfate the c

hemist has added to the flask. Round your answer to significant digits.

Chemistry
1 answer:
FrozenT [24]3 years ago
8 0

Answer:

The question is incomplete as some details are missing. Here is the complete question ; A chemist adds 45.0mL of a 0.434M copper(II) sulfate CuSO4 solution to a reaction flask. Calculate the mass in grams of copper(II) sulfate the chemist has added to the flask. Round your answer to 2 significant digits

Explanation:

The step by step explanation is as shown in the attachment

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which of the following elements are in this unbalanced chemical reaction: li(s) + H2O(l) lioh(aq)+h2(g)​
Arada [10]

Answer:

Li and H

Explanation:

2Li(s)+2H2O(i)→2LiOH(aq)+H2(g) is full balanced

5 0
2 years ago
Read 2 more answers
Find the quantinum numbers n,m,l,s for the last of potassium layer pleasee help explain correctly all
Fantom [35]

Answer:

Quantum numbers of the outermost electron in potassium:

  • n = 4.
  • l  = 1.
  • m_l = 0.
  • Either m_s = 1/2.

Explanation:

Refer to the electron configuration of a potassium atom. The outermost electron in a ground-state potassium atom is in the 4s orbital (fourth s orbital.)

The quantum number n (the principal quantum number) specifies the main energy shell of an electron. This electron is in the fourth main energy shell (as seen in the number four in the orbital.) Hence, n = 4 for this electron.

The quantum number l (the angular momentum quantum number) specifies the shape (s, p, d, etc.) of an electron. l = 1 for s\! orbitals (such as the one that contains this electron.

Quantum numbers n and l specify the shape of an orbital. On the other hand, the magnetic quantum number m_l specifies the orientation of these orbitals in space.

However, s orbitals are spherical. Regardless of the value of n, the only possible m_l value for electrons in s\! orbitals is m_l = 0.

The spin quantum number m_s distinguishes between the two electrons in an orbital. The two possible values of m_s \! are (+1/2) and (-1/2). Typically, the first electron in an orbital is assigned an upward (\uparrow) spin, which corresponds to m_s = (+1/2).

5 0
3 years ago
What’s the answers to these?
nata0808 [166]

Answer:

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Explanation:

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5 0
3 years ago
. an ionic compound is composed of 34.95 g of iron and 15.05 g of oxygen. find the empirical formula of this compound.
Contact [7]

Empirical formula of ionic compound is FeO. In which the composition of atoms is 1 : 1.

Empirical formula of an ionic compound is defined as the formula which gives whole number ratio of atoms of various elements present in molecule of compund.

mass of iron in compound = 34.95 g

molar mass of iron = 55.8 g

mass of oxygen in compound = 15.05 g

molar mass of oxygen = 32 g

number of moles of iron present in the compound are ratio of mass of iron in compound/ molar mass of iron

number of moles of iron in compound= 34.95 / 55.8 = 0.6263 ~ 1

number of moles oxygen in compound= 15.05/ 32 = 0.473 ~ 0.5

the ratio of the number of oxygen atoms to number of iron atoms present in one formula unit of iron compund is 2×0.5 / 1 = 1 : 1

Hence , the required empirical formula of iron compound is FeO.

To learn more about Emiprical formula, refer:

brainly.com/question/1439914

#SPJ4

3 0
1 year ago
Calculate the oxidation numbers of the elements in this equation<br> 2 Al2O3 → 4 Al + 3 O2
riadik2000 [5.3K]

Explanation:

Answer

Open in answr app

The rule used here is that the algebraic sum of the oxidation numbers of all the atoms a molecule is zero.

Al2O32× ( oxidation number of Al)+3× ( Oxidation number of O )  = 0

2× ( Oxidation number of Al) +3(−2)=0

2× ( oxidation number of Al) +6

∴ Oxidation number of Al =+3

5 0
3 years ago
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