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velikii [3]
3 years ago
9

What group would this element be in?

Chemistry
1 answer:
Lesechka [4]3 years ago
6 0
Non-metals. it is fluorine i believe and in the periodic table it is under the non-metal category
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What mass of calcium carbonate is produced when 250 mL of 6.0 M sodium carbonate is added to 750 mL of 1.0 M calcium fluoride
Savatey [412]

<u>Given:</u>

Volume of Na2CO3 = 250 ml = 0.250 L

Molarity of Na2CO3 = 6.0 M

Volume of CaF2 = 750 ml = 0.750 L

Molarity of CaF2 = 1.0 M

<u>To determine:</u>

The mass of CaCO3 produced

<u>Explanation:</u>

Na2CO3 + CaF2 → CaCO3 + 2NaF

Based on the reaction stoichiometry:

1 mole of Na2CO3 reacts with 1 moles of Caf2 to produce 1 mole of caco3

Moles of Na2CO3 present = V * M = 0.250 L * 6.0 moles/L = 1.5 moles

Moles of CaF2 present = V* M = 0.750 * 1 = 0.750 moles

CaF2 is the limiting reagent

Thus, # moles of CaCO3 produced = 0.750 moles

Molar mass of CaCO3 = 100 g/mol

Mass of CaCO3 produced = 0.750 moles * 100 g/mol  = 75 g

Ans: Mass of CaCO3 produced = 75 g

7 0
3 years ago
Plz help ugh!! will upvote
erma4kov [3.2K]
All are CORRECT except (d)
6 0
3 years ago
What can cause a mutation when the cell copies its DNA?
Sedaia [141]

Answer:

exposure to radiation i think

Explanation:

6 0
3 years ago
Read 2 more answers
Do parts a, b and c
Leona [35]

Answer:- (a)The pH of the buffer solution is 3.90.

(b) the pH of the solution after addition of HCl would be 3.60.

(c) the pH of the buffer solution after addition of NaOH is 4.32.

Solution:- (a) It is a buffer solution so the pH could easily be calculated using Handerson equation:

pH=Pka+log(\frac{base}{acid})

pKa can be calculated from given Ka value as:

pKa=-logKa

pKa=-log(6.3*10^-^5)

pKa = 4.20

let's plug in the values in the Handerson equation:

pH=4.20+log(\frac{0.025}{0.05})

pH = 4.20 - 0.30

pH = 3.90

The pH of the buffer solution is 3.90.

(b) Let's say the acid is represented by HA and the base is represented by A^- .

Original mili moles of HA from part a = 0.05(100) = 5

original mili moles of A^- from part a = 0.025(100) = 2.5

mili moles of HCl that is H^+ added = 0.100(10.0) = 1

This HCl reacts with the base present in the buffer to make HA as:

A^-+H^+\rightarrow HA

Total mili moles of HA after addition of HCl = 5+1 = 6

mili moles of base after addition of HCl = 2.5-1 = 1.5

Let's plug in the values in the Handerson equation again. Here, we could use the mili moles to calculate the pH. The answer remains same even if we use the concentrations also as the final volume is same both for acid and base.

pH=4.20+log(\frac{1.5}{6})

pH = 4.20 - 0.60

pH = 3.60

So, the pH of the solution after addition of HCl would be 3.60.

(c) mili moles of NaOH or OH^- added to the original buffer = 0.05(15.0) = 0.75

This OH^- reacts with HA to form A^- as:

HA+OH^-\rightarrow H_2O+A^-

mili moles of HA after addition of NaOH = 5-0.75 = 4.25

mili moles of A^- after addition of NaOH = 2.5+0.75 = 3.25

Let's plug in the values again in Handerson equation:

pH=4.20+log(\frac{3.25}{4.25})

pH = 4.20 - 0.12

pH = 4.32

So, the pH of the buffer solution after addition of NaOH is 4.32.

7 0
3 years ago
A newly discovered unicellular organism from acidic mine drainage is found to contain a cell wall, a plasma membrane, two flagel
makvit [3.9K]

Answer:

<em>The correct option is D)  a motile eukaryote</em>

Explanation:

Peroxisomes are structures which are present only in the plant or animal cells. Hence, based on this observation we can easily tell that the organism which is newly discovered is a eukaryote. Both plants and animals are eukaryotes.

Flagella are structures which are thin and hair shaped. Flagella allow movement of a cell, mostly they allow a whip-like movement. Based on this observation we will conclude that the organism discovered is a motile eukaryote.

5 0
3 years ago
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