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3 years ago

WHAT MEASURING EQUIPMENT DID LEANNE USE TO GET HER RESULTS

Chemistry

2 answers:

julia-pushkina [17]3 years ago

7 0

Answer:

Thietbi od

Explanation:

uranmaximum [27]3 years ago

5 0

Projector equator sargon

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Given 1.00 grams of each of the following radioisotopes, Ca-37, U-238, P-32 and Rn-222,which would have the least remaining orig

erica [24]

1) Ca-37, with a half-life of 181.1(10) ms.

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4 years ago

Measurements include

Keith_Richards [23]

There’s lots of measurements. (m, kg, s, mol, cm, in, mm) etc

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4 years ago

A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s

Alla [95]

Answer: The percent yield of the 1-bromobutane is 48.65 %

Explanation:

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

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

For NaBr:

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol$

The chemical equation for the reaction of 1-butanol and NaBr is:

$\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}$

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = $\frac{1}{1}\times 0.108=0.108$ moles of 1-bromobutane

Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

$0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g$

To calculate the percentage yield of 1-bromobutane, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

$\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%$

Hence, the percent yield of the 1-bromobutane is 48.65 %

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3 years ago

Yesterday we combined Hydrochloric Acid HCl with Sodium Hydroxide NaOH in a violent reaction that resulted in water H2O and comm

IceJOKER [234]

Reaction:
HCl + NaOH ---> NaCl + H2O

1 mole of HCl = 36,5 g
1 mole of NaOH = 40g

so, according to the reaction:
1 mol HCl = 1 mol NaOH
so, we need > 36,5 g HCl (hydrochloric acid)

answer: 36,5 g HCl (hydrochloric acid)
 ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
next question.

1 mole of NaCl = 58,5 g
1 mole of H2O = 18g

so, according to the reaction:
1 mole of HCl (36,5 g) ----------------- - 1 mole of NaCl (58,5 g)
(the same for NaOH)
i
1 mole of HCl (36,5 g) ------------------ 1 mole of H2O (18 g)
(the same for NaOH)

so, this reaction is stechiometric

answer: 58,5 g NaCl i 18g H2O

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4 years ago

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Plsss helpppp im confused will mark BRAINLIEST

mr_godi [17]

I am thinking Li & CI :)

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3 years ago

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