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aleksklad [387]
3 years ago
14

Scientists' belief that an atom is a solid sphere, is good enough to explain; PLZ I HAVE LESS TIME!!!!

Chemistry
1 answer:
Natasha2012 [34]3 years ago
7 0

Answer:

Its quite vague, instead you could say an atom is the smallest building block which further consists of subatomic particles like protons, neutrons and electrons :)

Hope thi helps :) and I'd appreciate if you'd mark brainliest because ive been stuck on the same rank for quite a long time :(

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The chemist discovers a more efficient catalyst that can produce ethyl butyrate with a 78.0% yield. How many grams would be prod
Vedmedyk [2.9K]

The grams that would be produced  from 7.70 g of butanoic acid and excess ethanol is 7.923grams

calculation

Step 1: write the chemical equation for the reaction

CH3CH2CH2COOH + CH3CH2OH →  CH3CH2CH2COOCH2CH3  +H2O

step 2: find the moles of butanoic acid

moles= mass/ molar mass

=  7.70 g/ 88 g/mol=0.0875 moles

Step 3:  use the mole ratio to determine the moles of ethyl butyrate

moles ratio of CH3CH2CH2COOH :CH3CH2CH2COOCH2CH3  is 1:1 therefore the moles of CH3CH2CH2COOCH2CH3 = 0.0875  x78/100=0.0683moles

step 4: find mass = moles x molar mass

 =  0.0683 moles  x116 g/mol=7.923grams

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Explanation:

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a water sample is found to have a cl- content of 100ppm as nacl what is the concentration of chloride in moles per liter
ladessa [460]

Answer:

The concentration of chloride ion is 2.82\times10^{-3}\;mol/L

Explanation:

We know that 1 ppm is equal to 1 mg/L.

So, the Cl^- content 100 ppm suggests the presence of 100 mg of Cl^- in 1 L of solution.

The molar mass of Cl^- is equal to the molar mass of Cl atom as the mass of the excess electron in Cl^- is negligible as compared to the mass of Cl atom.

So, the molar mass of Cl^- is 35.453 g/mol.

Number of moles = (Mass)/(Molar mass)

Hence, the number of moles (N) of Cl^- present in 100 mg (0.100 g) of Cl^- is calculated as shown below:

N=\frac{0.100\;g}{35.453\;g/mol}=2.82\times 10^{-3}\;mol

So, there is 2.82\times10^{-3}\;mol of Cl^- present in 1 L of solution.

5 0
3 years ago
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