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Shkiper50 [21]
3 years ago
11

What would happen if you took a normal shaped bottle, sailed it at 1,000 feet, and then carried it up to 14,000 feet

Chemistry
2 answers:
Dovator [93]3 years ago
4 0

it would go 15000 feet

alexandr402 [8]3 years ago
4 0
It will make 15,000 because if you add 14000 + 1,000 that equals 15000
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What is the formula of the chromium(iii) complex that contains two ammonia and four thiocyanate (scn−) ligands?
Nonamiya [84]
Answer is: formula of the complex is [Cr(NH₃)₂(SCN)₄<span>]</span>⁻<span>.
This complex has negative charge (-1) because chromium (central atom or metal) has oxidation number +3, first ligand ammonia has neutral charge and second ligand thiocyanate has negative oxidation number -1:
+3 + 2</span>·0 + 4·(-1) = -1.
4 0
3 years ago
If uranium-235 is bombarded with a neutron, it will temporarily become
stealth61 [152]
All atoms consist of a nucleus with protons and neutrons that is surrounded by a sea of electrons. If Uranium-235 is bombarded with a neutron then the atom's atomic number will remain the same but its mass will increase by 1 unit. Thus, it will temporarily become Uranium-236. 
6 0
3 years ago
Which of the following is true?
kotykmax [81]
Anwser is C for cesium
6 0
3 years ago
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What does the roman numeral stand for in copper(1) oxide should it not be copper(II) oxide
photoshop1234 [79]

Answer:

The roman numeral in copper(I) oxide indicates that the oxidation number of copper in the compound is 1.

Explanation:

Roman numeral is used to indicate the oxidation number of an element in a compound.

The roman numeral in copper(I) oxide indicates that the oxidation number of copper in the compound is 1.

This can be seen from the following illustration:

copper(I) oxide => Cu₂O

Oxidation number of O = –2

Oxidation number of Cu₂O = 0

Oxidation number of Cu =?

Cu₂O = 0

2Cu + O = 0

2Cu – 2 = 0

Collect like terms

2Cu = 0 + 2

2Cu = 2

Divide both side by 2

Cu = 2/2

Cu = 1

Thus, we can see that the oxidation number of Cu in Cu₂O is 1. Hence the name of Cu₂O is copper(I) oxide indicating that the oxidation number of of copper (Cu) in the compound is 1.

For copper(II) oxide, we shall determine the oxidation number of Cu. This can be obtained as follow:

copper(II) oxide, CuO => CuO

Oxidation number of O = –2

Oxidation number of CuO = 0

Oxidation number of Cu =?

CuO = 0

Cu + O = 0

Cu – 2 = 0

Collect like terms

Cu = 0 + 2

Cu = 2

Thus, the oxidation number of Cu in CuO is 2. Hence the name of CuO is copper(II) oxide indicating that the oxidation number of of copper (Cu) in the compound is 2.

From the above illustrations,

We can see that the roman numeral in both copper(I) oxide, Cu₂O and copper(II) oxide, CuO are different because the oxidation number of Cu in both cases are different.

3 0
3 years ago
Which solute would be more effective at lowering the freezing point of water: MgCl2 and KNO3? Explain.
Phantasy [73]

Answer:

AlCl₃.

Explanation:

Adding solute to water causes depression of the boiling point.

The depression in freezing point (ΔTf) can be calculated using the relation:

ΔTf = i.Kf.m,

where, ΔTf is the depression in freezing point.

i is the van 't Hoff factor.

van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

Kf is the molal depression constant of water.

m is the molality of the solution (m = 1.0 m, for all solutions).

(1) NaCl:

i for NaCl = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (NaCl) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(2) MgCl₂:

i for MgCl₂ = no. of particles produced when the substance is dissolved/no. of original particle = 3/1 = 3.

∴ ΔTb for (MgCl₂) = i.Kb.m = (3)(Kf)(1.0 m) = 3(Kf).

(3) NaCl:

i for KBr = no. of particles produced when the substance is dissolved/no. of original particle = 2/1 = 2.

∴ ΔTb for (KBr) = i.Kb.m = (2)(Kf)(1.0 m) = 2(Kf).

(4) AlCl₃:

i for AlCl₃ = no. of particles produced when the substance is dissolved/no. of original particle = 4/1 = 4.

∴ ΔTb for (CoCl₃) = i.Kb.m = (4)(Kf)(1.0 m) = 4(Kf).

So, the ionic compound will lower the freezing point the most is: AlCl₃

4 0
3 years ago
Read 2 more answers
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