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guapka [62]
3 years ago
5

Prove that circle p and circle q are similar.

Chemistry
1 answer:
abruzzese [7]3 years ago
8 0
All circles are similar.
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What is the balance equation for <br> _AI+_Pb(NO3)2—__Pb+__Ca(NO3)2
Firdavs [7]

Answer:

2 Al + 3 Pb(NO3)2 > 2 Al(NO3)3 + 3 Pb

Explanation:

4 0
3 years ago
Magnetite is a binary compound containing only iron and oxygen. the percent, by weight, of iron is 72.360%. what is the empirica
hram777 [196]
Answer is: <span>c. Fe</span>₃<span>O</span>₄<span>.
</span>ω(Fe) = 72,360%.
ω(O) = 100% - 72,36% = 27,64%.
For example, if we the mass of compound is 100 g:
m(Fe) = 72,36 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 72,36 g ÷ 55,85 g/mol.
n(Fe) = 1,296 mol.
n(O) = 27,64 g ÷ 16 g/mol.
n(O) = 1,727 mol.
n(Fe) : n(O) = 1,296 mol : 1,727 mol.
n(Fe) : n(O) = 1 : 1,33 or 3 : 4.
7 0
3 years ago
Read 2 more answers
Identify some other substances (besides KCl) that might give a positive test for chloride upon addition of AgNO3. do you think i
nlexa [21]

Answer:

-The other substances that give a positive test with AgNO3 are other chlorides present, iodides and bromide.

-It is reasonable to exclude iodides and bromides but it is not reasonable to exclude other chlorides

Explanation:

In the qualitative determination of halogen ions, silver nitrate solution(AgNO3) is usually used. Now, various halide ions will give various colours of precipitate when mixed with with silver nitrate. For example, chlorides(Cl-) normally yield a white precipitate, bromides(Br-) normally yield a cream precipitate while iodides (I-) normally yield a yellow precipitate. Thus, all these ions or some of them may be present in the system.

With that being said, if other chlorides are present, they will also yield a white precipitate just like KCl leading to a false positive test for KCl. However, since other halogen ions yield precipitates of different colours, they don't lead to a false test for KCl. Thus, we can exclude other halides from the tendency to give us a false positive test for KCl but not other chlorides.

5 0
3 years ago
Explain how hydrogen bonding contributes to water's high heat of vaporization
pashok25 [27]

The heat/enthalpy of vaporization of water represents the energy input required to convert one mole of water into vapor at a constant temperature. Intermolecular forces including hydrogen bondings of significant strength hold water molecules in place under its liquid state. Whereas the molecules experience almost no intermolecular interactions under the gaseous state- consider the way noble gases molecules interact. It is thus necessary to supply sufficient energy to overcome all intermolecular interactions present in the substance under its liquid state to convert the substance into a gas. The heat of vaporization is thus related to the strength of the intermolecular interactions.

Water molecules contain hydrogen atoms bonded directly to oxygen atoms. Oxygen atoms are highly electronegative and take major control of electrons in hydrogen-oxygen bonds. Hydrogen atoms in water molecules thus experience a strong partial-positive charge and would attract lone pairs of electron on neighboring water molecules. "Hydrogen bonds" refer to the attraction between hydrogen atoms bonded to electronegative elements and lone pairs of electrons. The hydrogen-oxygen bonds in water molecules are so polarized that hydrogen bonds in water are stronger than both dipole-dipole interactions and London Dispersion Forces in most other molecules. It thus take high amounts of energy to separate water molecules sufficiently apart such that they no longer experience intermolecular interactions and behave collectively like a gas. As a result, water has one of the highest heat of vaporization among covalent molecules of similar sizes.

4 0
3 years ago
Solve the 3 questions
Dmitrij [34]

Answer:

what kind of math is this

Explanation:

5 0
3 years ago
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