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bija089 [108]
3 years ago
15

the size of the negative charge of an electron us exactly the same as the size of the positive charge of a proton. what is the o

verall charge of the helium atom ​
Chemistry
1 answer:
Ket [755]3 years ago
5 0
Surrounded by a cloud of negatively charged electrons The atomic nucleus consists of positively charged protons and electrically neutral neutrons. (except in the case of Hydrogen-1, which is
154 KB (12,134 words) - 15:57, April 9, 2021
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Metals are easy to bend beacuse
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Since metals are malleable they are able to be bent and shaped.
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4 years ago
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Please Help! Thank You
Ivanshal [37]

Explanation:

the answer is true because I had this question and got it right

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3 years ago
Discuss whether two products are required for a chemical reaction. (NEED HELP ASAP)
Alex787 [66]

Answer:

Yes.

Explanation:

"These starting substances of a chemical reaction are called the reactants, and the new substances that result are called the products."

There is a beginning product, and a reactant is needed in order for something to happen.

For example, according to Newton, something cannot happen until an exterior force comes and stops/pushes it.

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6 0
3 years ago
What is the name of the element with a valence electron configuration of 4s24p3?
Gnesinka [82]
<span>The element Arsenic, or As in the periodic table, possesses the valence electron configuration of 4s2 4p3. It has an atomic number of 33 and an atomic mass of 74.92, and is classified as a metalloid.</span>
4 0
4 years ago
6NaBr+1AlO3=3Na2O+2AlBr3 How many grams of NaBr would be needed in order to make 23.5 grams of AlBr3
Evgesh-ka [11]

Answer:

23.5 grams of AlBr3 will be produced by 27.20 grams of NaBr

Explanation:

The balanced equation here is

6NaBr + 1AlO3 = 3Na2O + 2AlBr3

6 moles of NaBr are required to produce 2 moles of AlBr3

Mass of one mole of NaBr = 102.894 g/mol

Mass of one mole of AlBr3 = 266.69 g/mol

Mass of 6 moles of NaBr = 6*102.894 g/mol

Mass of two moles of AlBr3 = 2*266.69 g/mol

6*102.894 g  NaBr produces 2*266.69 g of AlBr3

23.5 grams of AlBr3 will be produced by (6*102.894)/(2*266.69 )*23.5 = 27.20 grams of NaBr

6 0
3 years ago
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