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2 years ago

What is the result of rock’s poor ability to transmit heat?

Chemistry

1 answer:

Anastaziya [24]2 years ago

7 0

It is a thermal insulator because it doesn’t transfer heat.

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Is The reaction time is different for different stimuli true or false?

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Answer: true

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3 years ago

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2 years ago

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Explain why the several different types of microscopes are all necessary.

givi [52]

The different types of microscopes are all necessary because not all experiments require the same level of magnification. For dissections low magnification is sufficient, so a dissecting microscope works very well, while for viewing single cells the 1000 fold magnification of a compound light microscope is more accurate.

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3 years ago

Ephedrine, a central nervous system stimulant, is used in nasalsprays as a decongestant. this compound is a weak organic base: {

sashaice [31]

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base:

C10H15ON (aq) + H2O (l) -> C10H15ONH+ (aq) + OH- (aq)

A 0.035 M solution of ephedrine has a pH of 11.33.

a) What are the equilibrium concentrations of C10H15ON, C10H15ONH+, and OH-?

b) Calculate Kb for ephedrine.

c(C₁₀H₁₅NO) = 0,035 M.
pH = 11,33.
pOH = 14 - 11,33 = 2,67.
[OH⁻] = 10∧(-2,67) = 0,00213 M.
[OH⁻] = [C₁₀H₁₅NOH⁺] = 0,00213 M.
[C₁₀H₁₅NO] = 0,035 M - 0,00213 M = 0,03287 M.
Kb = [OH⁻] · [C₁₀H₁₅NOH⁺] / [C₁₀H₁₅NO].
Kb = (0,00213 M)² / 0,03287 M = 1,38·10⁻⁴.

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3 years ago

Hydrogen iodide, HI, is formed in an equilibrium reaction when gaseous hydrogen and iodine gas are heated together. If 20.0 g of

Kaylis [27]

Answer: D. 19.9 g hydrogen remains.

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $H_2$

$\text{Number of moles}=\frac{20.0g}{2g/mol}=10.0moles$

b) moles of $I_2$

$\text{Number of moles}=\frac{20.0g}{254g/mol}=0.0787moles$

$H_2(g)+I_2(g)\rightarrow 2HI(g)$

According to stoichiometry :

1 mole of $I_2$ require 1 mole of $H_2$

Thus 0.0787 moles of $l_2$ require= $\frac{1}{1}\times 0.0787=0.0787moles$ of $H_2$

Thus $l_2$ is the limiting reagent as it limits the formation of product and $H_2$ acts as the excess reagent. (10.0-0.0787)= 9.92 moles of $H_2$ are left unreacted.

Mass of $H_2=moles\times {\text {Molar mass}}=9.92moles\times 2.01g/mol=19.9g$

Thus 19.9 g of $H_2$ remains unreacted.

5 0

2 years ago