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3 years ago

Limescale removers contain sulfamic acid, H3NSO3, which reacts with minerals in hard water

Chemistry

1 answer:

I am Lyosha [343]3 years ago

4 0

Answer:

0.375 moles of CaCO₃ are required

Explanation:

Given data:

Number of moles of sulfamic acid = 0.75 mol

Number of moles of calcium carbonate required = ?

Solution:

Chemical equation:

2H₃NSO₃ + CaCO₃ → Ca(SO₃NH₂)₂ + CO₂ + H₂O

Now we will compare the moles of H₃NSO₃ and CaCO₃ .

H₃NSO₃ : CaCO₃

2 : 1

0.75 : 1/2×0.75 = 0.375 mol

Thus, 0.375 moles of CaCO₃ are required.

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Group 1 elements have an average electronegativity of 0.84 (not including hydrogen). Group 17 elements have an average electrone

Ksenya-84 [330]

The elements of group 1 makes ionic bond with the elements of group 7 due to high difference of electronegativity values.

<h3>Type of bond between group 1 and 7</h3>

The elements of first group lose its one outermost electrons while on the other hand, the elements of seven group needs one electron so they gain that one electron so they make an ionic bond with each other.

So we can conclude that the elements of group 1 makes ionic bond with the elements of group 7 because of the high difference of electronegativity values.

Learn more about electronegativity here: brainly.com/question/2415812

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2 years ago

Bonds between two atoms that are equally electronegative are _____.

statuscvo [17]

Bonds between two atoms that are equally electronegative are nonpolar covalent bonds.

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3 years ago

What is the correct Lewis structure for group 5A element arsenic?

stellarik [79]

I believe the anwer is A

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4 years ago

Read 2 more answers

What volume (mL) of the partially neutralized stomach acid was neutralized by NaOH during the titration? (portion of 25.00 mL sa

almond37 [142]

The question is incomplete, here is the complete question:

What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)

<u>Answer:</u> The volume of HCl neutralized is 1.25 mL

<u>Explanation:</u>

To calculate the volume of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of stomach acid which is HCl

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=1\\M_1=2M\\V_1=?mL\\n_2=1\\M_2=0.1M\\V_2=25mL$

Putting values in above equation, we get:

$1\times 2\times V_1=1\times 0.1\times 25\\\\V_1=\frac{1\times 0.1\times 25}{1\times 2}=1.25mL$

Hence, the volume of HCl neutralized is 1.25 mL

3 0

3 years ago

One of the nuclides in each of the following pairs is radioactive. Predict which is radioactive and which is stable.a. 39/19K an

marishachu [46]

Answer:

a) 39/19 K : stable nuclide, 40/19 K : radioactive nuclide.

b) 209B: stable nuclide, 208Bi : radioactive nuclide

c) nickel-58 : stable nuclide, nickel-65 : radioactive nuclide.

Explanation:

As per the rule, nuclides having odd number of neutrons are generally not stable and therefore, are radioactive.

Mass number (A) = Atomic number (Z) + No. of neutrons (N)

Or, N = A - Z

39/19 K and 40/19 K

Calculate no. of neutrons in 39/19 K as follows:

atomic no. = 19, mass no. 39

N = 39 - 19

= 20 (even no.)

Calculate no. of neutrons in 40/19 K as follows:

atomic no. = 19, mass no. 40

N = 40 - 19

= 21 (odd no.)

Therefore, 39/19 K is a stable nuclide and 40/19 K is a radioactive

nuclide.

209Bi and 208Bi

Calculate no. of neutrons in 209Bi as follows:

atomic no. = 83, mass no. 209

N = 209 - 83

= 126 (even no.)

Calculate no. of neutrons in 208Bi as follows:

atomic no. = 83, mass no. 208

N = 208 - 83

= 125 (odd no.)

Therefore, 209Bi is a stable nuclide and 208Bi is a radioactive nuclide.

nickel-58 and nickel-65

Calculate no. of neutrons in nickel-58 as follows:

atomic no. = 28, mass no. 58

N = 58 - 28

= 30 (even no.)

Calculate no. of neutrons in nickel as follows:

atomic no. = 28, mass no. 65

N = 65 - 28

= 37 (odd no.)

Therefore,nickel-58 is a stable nuclide and nickel-65 is a radioactive nuclide.

5 0

3 years ago