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Art [367]
3 years ago
7

Which of the following elements is most likely to form a covalent bond?

Chemistry
1 answer:
Alecsey [184]3 years ago
7 0
Bromine, it's the only element that isn't a metal
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Should we only write condensed formula in chemical equations (for lessons like carbon and compounds). What did your teachers say
Firlakuza [10]

Answer:

Here's what I get.

Explanation:

  • If your teachers don't ask for a specific type of formula, a condensed structural formula should be OK.
  • If they ask specifically for a structural formula or a bond-line formula, that is what you must give.

Bottom line: ask your teachers in advance what they expect.

7 0
3 years ago
You have 180 g of radium that takes up 36 cm ^ 3 of space if you were to cut it in half and have only 90 g
Leno4ka [110]

If you were to cut the radium  in half and have only 90 g, it will take up 18 cm³.

<h3>What is density?</h3>

The density of an object is the ratio of mass to volume of object.

Density = mass/volume

volume = mass/density

at a constant density, the volume of an object is proportional to its mass.

From the question, you have 180 g of radium that takes up 36 cm ^ 3 of space if you were to cut it in half and have only 90 g, the new mass will take the following volume.

180 g = 36 cm³

90 g =  ?

= (90 x 36) / 180

= 18 cm³

Thus, if you were to cut the radium  in half and have only 90 g, it will take up 18 cm³.

Learn more about radium here: brainly.com/question/23781489

#SPJ1

6 0
1 year ago
I think its A can someone help me?
ratelena [41]
The answer is A) Scientific Law
5 0
3 years ago
Read 2 more answers
Calculate the pH of 1.00 L of a buffer that contains 0.105 M HNO2 and 0.170 M NaNO2. What is the pH of the same buffer after the
kari74 [83]

Answer:

1.- pH =3.61

2.-pH =3.53

Explanation:

In the first part of this problem we can compute the pH of the buffer by making use of the Henderson-Hasselbach equation,

pH = pKa + log [A⁻]/[HA]

where [A⁻] is the conjugate base anion concentration ( [NO₂⁻]), [HA] is the weak acid concentration,[HNO₂].

In the second part, our strategy has to take into account that some of the weak base NO₂⁻ will be consumed by reaction with the very strong acid HCl. Thus, first we will calculate the new concentrations, and then find the new pH similar to the first part.

First Part

pH  = 3.40+ log {0.170 /0.105}

pH =  3.61

Second Part

# mol HCl = ( 0.001 L ) x 12.0 mol / L = 0.012

# mol NaNO₂ reacted = 0.012 mol ( 1: 1 reaction)

# mol NaNO₂ initial = 0.170 mol/L x 1 L = 0.170 mol

# mol NaNO₂ remaining = (0.170 - 0.012) mol  = 0.158

# mol HNO₂ produced = 0.012 mol

# mol HNO₂ initial = 0.105

# new mol HNO₂ = (0.105 + 0.012) mol = 0.117 mol

Now we are ready to use the Henderson-Hasselbach with the new ration. Notice that we dont have to calculate the concentration (M) since we are using a ratio.

pH = 3.40 + log {0.158/.0117}

pH = 3.53

Notice there is little variation in the pH of the buffer. That is the usefulness of buffers.

4 0
2 years ago
Read 2 more answers
A certain alkyl halide is reacted with OH- to form an alcohol. The alkyl halide is optically active but the product(s) is/are op
Snowcat [4.5K]

Answer:

a. 3-brumo - 3-methylhexane

Explanation:

Alkyl Halides can undergo substitution reactions. Nucleophiles are electron rich species and has negative charge while Electrophiles are electron deficient species which carry positive charge. Alkyl halide which have polar carbon atom are electrophiles.

8 0
3 years ago
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