All categories

3 years ago

2.1 x 10^-3 x 2 x 10^2 =

Chemistry

1 answer:

Bumek [7]3 years ago

4 0

Answer:

I believe it is B

Explanation:

hope it helps. please let me know if it's wrong

You might be interested in

What mass of chromium would be produced from the reaction of 57.0 g of potassium with 199 g of chromium(II) bromide according to

fredd [130]

Answer:

Mass of Chromium produced = 37.91 grams

Explanation:

2K + CrBr₂ → 2KBr + Cr

2mole 1 mole 1 mole

mass of Potassium = 57.0 grams

molar mass of Potassium = 39.1 g/mol

no of moles of Potassium = 57.0 / 39.1 = 1.458 moles

mass of CrBr₂= 199 grams

molar mass of CrBr₂ = 211.8 gram/mole

no of moles of CrBr₂ = 199 / 211.8 = 0.939 mole

From chemical equation

1 mole of CrBr₂ = 2 moles of K

∴ 0.939 moles of CrBr₂ = ?

⇒ 0.939 x 2/1 = 1.878 moles of K

1.878 moles of K is needed, but there is 1.458 moles of K. So, Potassium is completed first during the reaction . Hence, Potassium is limiting reagent. and CrBr₂ is excess reagent .

From chemical equation

2 moles of K = 1 mole of Cr

∴ 1.458 moles of K = ?

⇒ 1.458 x 1/ 2 = 0.729 moles of Cr

no of moles of Cr formed = 0.729 moles

molar mass of Cr = 52.0 g/mol

mass of one mole of Cr = 52.0 grams

mass of 0.729 moles of Cr = 52.0 x 0.729 = 37.908 grams

mass of Chromium produced = 37.91 grams

6 0

3 years ago

A precipitate of lead(II) chromate can be formed using the reaction below.

Marina86 [1]

Answer:

Explanation:

7 0

3 years ago

Enter your answer in the provided box. Find the pH of a buffer that consists of 0.34 M HBrO and 0.89 M KBrO (PK, of HBrO = 8.64)

11111nata11111 [884]

Answer : The pH of buffer is 9.06.

Explanation : Given,

$pK_a=8.64$

Concentration of HBrO = 0.34 M

Concentration of KBrO = 0.89 M

Now we have to calculate the pH of buffer.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[KBrO]}{[HBrO]}$

Now put all the given values in this expression, we get:

$pH=8.64+\log (\frac{0.89}{0.34})$

$pH=9.06$

Therefore, the pH of buffer is 9.06.

6 0

3 years ago

How many moles are in 0.821g of lead?

Ivenika [448]

Answer:

3.96 × 10⁻³ mol Pb

Explanation:

Step 1: Find molar mass

Pb (Lead) - 207.2 g/mol

Step 2: Use Dimensional Analysis

$0.821 \hspace{2} g \hspace{2} Pb(\frac{1 \hspace{2} mol \hspace{2} Pb}{207.2 \hspace{2} g \hspace{2} Pb} )$ = 0.003962 mol Pb

Step 3: Simplify

We have 3 sig figs

0.003962 mol Pb ≈ 0.00396 mol Pb

0.00396 mol Pb = 3.96 × 10⁻³ mol Pb

3 0

4 years ago

Why does Napatha have a lower boiling point than disel?

Nostrana [21]

Boiling points are related to the type of intermolecular forces between the compounds. if napatha has a lower boiling point hen it must have weaker bonds in it. Stronger bonds equal higher boiling points

3 0

3 years ago