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Damm [24]
3 years ago
12

Help me please I really need it

Chemistry
1 answer:
AleksandrR [38]3 years ago
6 0
I think the answer is white.
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When a surface is experiencing friction with another surface, how are the particles
Free_Kalibri [48]

Answer:

the surface may become etheir scratchy and soft and that the surface may also be come heated up due to friction

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All cartilage ossifies. True or false <br><br><br><br> will mark brainliest!
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3 years ago
Select the correct answer.
Serjik [45]

Ionic bond is a chemical bond formed by the complete transfer of electrons between two atoms. The atom that loses electrons gains a positive charge (cation) and that which accepts electrons gains a negative charge (anion). Now, electronegativity is a parameter that measures the tendency of an atom to accept electrons. In the context of ionic bonding, two elements which show a significant difference in their electronegativity values form ionic bonds.

In the given examples, the difference in electronegativity is greatest between K  and Br i.e. 0.8 and 2.8 respectively with a difference of 2.0. This also makes sense since K and Br are on the extreme ends of the periodic table. Hence, potassium with a valence electron configuration of 4s1 will lose its s electron to Br (4s24p6) and form an ionic molecule K⁺Br⁻

Ans E) potassium and bromine

8 0
4 years ago
Read 2 more answers
Which of the following statement is not true about cooking techniques? 
nika2105 [10]

Answer:

b) Every experienced cooks at home tend not to use recipes, instead based entirely upon their knowledge of cooking technique.

Explanation:

This is because even the most experienced cooks might make a mistake, so having a recipe where they can follow is better than just use their knowledge. If his or her knowledge has a mistake it might create a big consequence.

8 0
3 years ago
How many grams of Br are in 335g of CaBr2
ASHA 777 [7]
The answer is 267.93 g

Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol

The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%

Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
7 0
3 years ago
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