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2 years ago

15

Which type of plate boundary is associated with seafloor spreading at the source of the spread: convergent, divergent, or transf

orm?
HURRY!

1 answer:

aev [14]2 years ago

8 0

The answer is divergent

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Define mechanical energy. Using the example of the pendulum, explain how kinetic and potential energy relate to the mechanical e

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Mechanical energy is the sum of potential energy and kinetic energy present in the components of a mechanical system. It is the energy associated with the motion and position of an object. All energy can be expressed in Joules (including thermal <span>energy</span>

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3 years ago

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Answer:

A

Explanation:

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A weak base is added to water. Which information is required to find the strength of its conjugate acid?

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Answer: C) the values of Kb and Kw

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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]

<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %

<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 NaBr:</u>

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

The specific heat of liquid ethyl alcohol is 2.42 j/g C and its density of 0.7893 g/mL. A piece of solid sliver (specific heat 0

Doss [256]

MAg*cAg*(T1-T)=ρalc*Valc*calc*(T-T2)
mAg=?(g)
cAg=0.24J/gC
T1=95
T=23.5
Valc=25.6ml
ρalc=0.7893g/ml
T2=19.27
use wollframalpha or calculator

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3 years ago

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