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3 years ago

Na+ FeBr2 what are the products

Chemistry

1 answer:

densk [106]3 years ago

8 0

Hello:

Balanced equation:

2 Na + FeBr2 = 2 NaBr + Fe

Reaction type: single replacement

Hope that helps!

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015

kozerog [31]

Answer:

Average of measurements = 20.97 cm

Explanation:

Given data:

Three measurements of pete's = 20.9 cm, 21.0 cm, 21.0 cm.

Average value of measurements = ?

Formula:

Average of measurements = sum of all measurements / Total number of measurements

Solution:

Average of measurements = 20.9 cm + 21.0 cm + 21.0 cm. / 3

Average of measurements = 62.9 cm / 3

Average of measurements = 20.97 cm

7 0

2 years ago

Ca + HCl -> CaCl2 + H2 Li + H2O -> LiOH + H2 NH3 + O2 -> NO + H2O K + O2 -> K2O

jonny [76]

Answer:

Are you trolling us?

Explanation:

5 0

3 years ago

Read 2 more answers

How many Helium atoms are in 12 grams of Helium?

horsena [70]

12gHe/1 × 1molHe/4.0026g × 6.02x10^23atomHe/1mol = 1.8 atoms

6 0

3 years ago

What is the relationship between frequency and wavelength?

IrinaVladis [17]

Answer:

frequency it the measure of the wave length. The measure of the peaks and troughs is how you measure the frequency. the distance between these is the wave length.

6 0

3 years ago

Assuming the volumes are additive, what is the [Cl−] in a solution obtained by mixing 297 mL of 0.675 M KCl and 664 mL of 0.338

Elden [556K]

Answer: The concentration of chloride ions in the solution obtained is 0.674 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For KCl:

Molarity of KCl solution = 0.675 M

Volume of solution = 297 mL

Putting values in equation 1, we get:

$0.675=\frac{\text{Moles of KCl}\times 1000}{297}\\\\\text{Moles of KCl}=\frac{(0.675mol/L\times 297)}{1000}=0.200mol$

1 mole of KCl produces 1 mole of chloride ions and 1 mole of potassium ion

Moles of chloride ions in KCl = 0.200 moles

For magnesium chloride:

Molarity of magnesium chloride solution = 0.338 M

Volume of solution = 664 mL

Putting values in equation 1, we get:

$0.338=\frac{\text{Moles of }MgCl_2\times 1000}{664}\\\\\text{Moles of }MgCl_2=\frac{(0.338mol/L\times 664)}{1000}=0.224mol$

1 mole of magnesium chloride produces 2 moles of chloride ions and 1 mole of magnesium ion

Moles of chloride ions in magnesium chloride = $(2\times 0.224)=0.448mol$

Calculating the chloride ion concentration, we use equation 1:

Total moles of chloride ions in the solution = (0.200 + 0.448) moles = 0.648 moles

Total volume of the solution = (297 + 664) mL = 961 mL

Putting values in equation 1, we get:

$\text{Concentration of chloride ions}=\frac{0.648mol\times 1000}{961}\\\\\text{Concentration of chloride ions}=0.674M$

Hence, the concentration of chloride ions in the solution obtained is 0.674 M

5 0

3 years ago