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2 years ago

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What is the chemical formula for Sodium bicarbonate

1 answer:

Cloud [144]2 years ago

5 0

Answer: Na2CO3

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Potassium is isotopic and has RAM of 39.5 work out the percentage abundance of each isotope in a given sample of potassium which

Troyanec [42]

Yo sup??

Let the percentage of K-39 be x

then the percentage of K-40 is 100-(x+0.01)

We know that the net weight should be 39.5. Therefore we can say

(39*x+40*(100-(x+0.01))+38*0.01)/100=39.5

(since we are taking it in percent)

39*x+40*(100-(x+0.01))+38*0.01=3950

39x+4000-40x-0.4+0.38=3950

2x=49.98

x=24.99

=25 (approx)

Therefore K-39 is 25% in nature and K-40 is 75% in nature.

Hope this helps.

3 0

3 years ago

Is zach gay and a joe biden

shtirl [24]

Answer:

yes they're gay

5 0

2 years ago

How many moles of c6h12o6 are consumed if 6 moles of o2 are consumed

Mrac [35]

C₆H₁₂O₆, or glucose, is oxidized in the presence of oxygen to form carbon dioxide and water. The reaction equation for this is:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O
Therefore, if 6 moles of oxygen are consumed, we can see from the equation that one mole of C₆H₁₂O₆ will be consumed.

3 0

3 years ago

A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out

scZoUnD [109]

<u>Answer:</u> The mass percent of potassium bromide in the mixture is 9.996%

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For lead (II) bromide:</u>

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

$\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol$

The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

$2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3$

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = $\frac{2}{1}\times 0.0021=0.0042mol$ of potassium bromide

Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

$0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g$

To calculate the percentage composition of KBr in the mixture, we use the equation:

$\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100$

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

$\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%$

Hence, the percent by mass of KBr in the mixture is 9.996 %

5 0

3 years ago

Water is believed to be one of the best fire extinguishers then why it is so that fire in oil spills over ocean/sea is difficult

Eddi Din [679]

Answer:

D

Explanation:

Trust me

6 0

3 years ago