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4 years ago

In the periodic table how many elements could be classified as metals in group IV?

Chemistry

1 answer:

lord [1]4 years ago

8 0

4 elements are classified in group 4 like titanium, zirconium, hafnium and ruthorfordium

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How does the rate of reaction proceed with time

Studentka2010 [4]

It is the speed of light

6 0

3 years ago

Calculate the molality of a 20.0 percent by weight aqueous solution of nh4cl. (molecular weight: nh4cl = 53.5)

cestrela7 [59]

Unfortunately the data provided doesn't include the DENSITY of the ammonium chloride solution and molarity is defined as moles per volume. So without the density, the calculation of the molarity is impossible. But fortunately, there are tables available that do provide the required density and for a 20% solution by weight, the density of the solution is 1.057 g/ml.
So 1 liter of solution will mass 1057 grams and the mass of ammonium chloride will be 0.2 * 1057 g = 211.4 g. The number of moles will then be 211.4 g / 53.5 g/mol = 3.951401869 mol. Rounding to 3 significant digits gives a molarity of 3.95.
Now assuming that your teacher wants you to assume that the solution masses 1.00 g/ml, then the mass of ammonium chloride will only be 200g, and that is only (200/53.5) = 3.74 moles.
So in conclusion, the expected answer is 3.74 M, although the correct answer using missing information is 3.95 M.

8 0

4 years ago

Read 2 more answers

Determine the concentration of H+ ions in an aqueous solution where [OH-] =2.25 × 10^-4 M. I have 29 minutes I beg u guys pls hu

mixas84 [53]

but where Is the volume in order for us to determine the concentration. since we have moles in H+ ions

then you can say

concentration = M*1000/V

6 0

3 years ago

200.0 mL of 3.85 M HCl is added to 100.0 mL of 4.6 M barium hydroxide. The reaction goes to completion. What is the concentratio

Ede4ka [16]

Answer:

2.387 mol/L

Explanation:

The reaction that takes place is:

2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

First we calculate how many moles of each reagent were added:

HCl ⇒ 200.0 mL * 3.85 M = 203.85 mmol HCl

Ba(OH)₂ ⇒ 100.0 mL * 4.6 M = 460 mmol Ba(OH)₂

460 mmol of Ba(OH)₂ would react completely with (2*460) 920 mmol of HCl. There are not as many mmoles of HCl so Ba(OH)₂ will remain in excess.

Now we calculate how many moles of Ba(OH)₂ reacted, by converting the total number of HCl moles to Ba(OH)₂ moles:

203.85 mmol HCl * $\frac{1mmolBa(OH)_{2}}{2mmolHCl}$ = 101.925 mmol Ba(OH)₂

This means the remaining Ba(OH)₂ is:

460 mmol - 101.925 mmol = 358.075 mmoles Ba(OH)₂

There are two OH⁻ moles per Ba(OH)₂ mol:

OH⁻ moles = 2 * 358.075 = 716.15 mmol OH⁻

Finally we divide the number of OH⁻ moles by the total volume (100 mL + 200 mL):

716.15 mmol OH⁻ / 300.0 mL = 2.387 M

So the answer is 2.387 mol/L

7 0

3 years ago

Please please !!! help me I’ve been doing this for 2 hours

MatroZZZ [7]

Answer:

3.7 x 10^23

is the right answer

6 0

3 years ago