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3 years ago

An element occurs naturally as a gas. It is also characterized by a very

Chemistry

1 answer:

Delvig [45]3 years ago

8 0

I believe it would be c

Please brainiest

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Clo3− draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons. Show t

Karo-lina-s [1.5K]

The formal charges of all nonhydrogen atoms are -1.

Solution:-

O 7-4 = 3 O Double bond on one H 5-4 = 1

O-Cl-O 6-7 = -1x4 = -4 N 5-4=1 H-N-H 1-1=0

O 3-4= -1 O O 6-7 = -1(2)=-2 H 1-0=+1

6-6 = 0 1-2 = -1

It will percentage its last valence electron thru a single bond to the terminal oxygen atom. This is in agreement with carbon and hydrogen atoms that each need to form 4 and 1 covalent bonds respectively. because the terminal oxygen atom best has a single covalent bond, it'll have a proper rate of -1.

According to the lewis structure of SO2, The critical atom is sulfur and it is bonded with 2 oxygen atoms thru a double bond. each oxygen atom acquires 2 lone pairs of electrons and the primary sulfur atom has 1 lone pair of electrons.

Learn more about Nonhydrogen atoms here:-brainly.com/question/2822744

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3 0

1 year ago

What is the expected mass (in kg) of a 10 over 5 B isotope? 1 proton = 1.6726 × 10-27 kg 1 neutron = 1.6749 × 10-27 kg A.1.67 ×

Ira Lisetskai [31]

$^{10}_5 B$
An atom of this isotope contains 5 protons and 10-5=5 neutrons.

$5 \times 1.6726 \times 10^{-27} + 5 \times 1.6749 \times 10^{-27} = \\
8.363 \times 10^{-27} + 8.3745 \times 10^{-27}= \\
16.7375 \times 10^{-27}= \\
1.67375 \times 10^{-26} \approx \\
1.67 \times 10^{-26}$

The answer is A. 1.67 × 10⁻²⁶ kg.

8 0

3 years ago

Select the one that most likely has the highest rate of effusion Oxygen (O2) or hydrogen (H2):

Serhud [2]

The answer is hydrogen (H2) because i just answered this question from Edgenuit.

8 0

3 years ago

Please help, I really don’t understand this!!!

kap26 [50]

Analysing the Question:

We are given the balanced equation:

C₆H₁₂O₆ + 6O₂→ 6CO₂ + 6H₂O

from this equation, we can say that: for every 1 mole of Glucose, we need 6 moles of Oxygen

Moles of Glucose used in the reaction:

Molar mass of Glucose = 180 grams / mol

Given mass of Glucose = 1 gram

Mole of Glucose = Given mass / Molar mass

Moles of Glucose = 1 / 180 moles

Mass of Oxygen required:

We know that for every mole of Glucose, we need 6 moles of Oxygen

So, for 1/180 moles of Glucose, we need 6 / 180 = 1 / 30 moles of Oxygen

Mass of 1 / 30 moles of Oxygen:

Mass = Molar mass * number of moles

Mass of Oxygen = 32 * 1/30

Mass of Oxygen = 32 / 30

Mass of Oxygen = 1.06 grams

5 0

3 years ago

Name the method that used to separated amino acids from fruit juice solution

Ket [755]

Using the method chromatography

8 0

3 years ago