What is the most reactive element on the periodic table

2 questions<br> Popeyes or KFC<br> Mcdonalds, Wendys, Burger King, Shake Shack, or Chick Fil A

Ymorist [56]

Answer:

Chick-fli-a

Explanation:

3 0

3 years ago

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The public is not yet able to purchase cars powered by hydrogen fuel cells because engineers have to determine how the cars perf

Leya [2.2K]

Answer:

It’s B just did it

Explanation:

6 0

3 years ago

Chemistry chemistry chemistry chemistry chemistry chemistry

fgiga [73]

Explanation:

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8 0

3 years ago

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Urea, (NH2)2CO, is a product of metabolism of proteins. An aqueous solution is 37.2% urea by mass and has a density of 1.032 g/m

Feliz [49]

Answer:

The molarity of urea in this solution is 6.39 M.

Explanation:

Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is

$molarity = moles of solute ÷ liters of solution$

To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.

Our first step is to calculate the moles of urea in 100 grams of the solution,

using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is

60.06 g/mol ÷ 37.2 g = 0.619 mol

Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.

1.032 g/mL ÷ 100 g = 96.9 mL

This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.

0.619 mol/96.9 mL × 1000 mL= 6.39 M

Therefore, the molarity of the solution is 6.39 M.

4 0

3 years ago

Let us write the appropriate equilibria and associate the correction <img src="https://tex.z-dn.net/?f=K_b" id="TexFormula1" tit

rosijanka [135]

Explanation:

The relation between $K_a\&K_b$ is given by :

$K_w=K_a\times K_b$

Where :

$K_w=1\times 10^{-14}$ = Ionic prodcut of water

The value of the first ionization constant of sodium sulfite = $K_{a1}=1.4\times 10^{-2}$

The value of $K_{b1}$ :

$1\times 10^{-14}=1.4\times 10^{-2}\times K_{b1}$

$K_{b1}=\frac{1\times 10^{-14}}{1.4\times 10^{-2}}=7.1\times 10^{-13}$

The value of the second ionization constant of sodium sulfite = $K_{a2}=6.3\times 10^{-8}$

The value of $K_{b2}$ :

$1\times 10^{-14}=6.3\times 10^{-8}\times K_{b1}$

$K_{b1}=\frac{1\times 10^{-14}}{6.3\times 10^{-8}}=1.6\times 10^{-7}$

3 0

3 years ago