3 years ago

Both kelp and diatoms are classified

Chemistry

1 answer:

Oduvanchick [21]3 years ago

6 0

They're both classified as <span>heterokonts

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The given formula for heat, Q=mc(Tf-Ti), is the best way to solve such problems with changes in temperature. It can be said that m is the mass of the substance. C is the specific heat of the substance. The term (Tf-Ti) is the change in temperature.

Q = mc(Tf-Ti) = 480g(0.96 J/g-C)(234-22) = 97689.6 Joules of heat

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3 years ago

Determine the expected diffraction angle for the first-order diffraction from the (111) set of planes for FCC nickel (Ni) when m

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Answer:

56°

Explanation:

First calculate $a:$

$a=2 R \sqrt{2}=2(0.1246) \sqrt{2}=0.352 \mathrm{nm}$

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$d_{111}=\frac{a}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{0.352}{\sqrt{3}}=0.203 \mathrm{nm}$

The diffraction angle is determined from:

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Solve for $\theta$

$\theta=\sin ^{-1}(0.476)=28^{\circ}$

The diffraction angle is:

$2 \theta=2\left(28^{\circ}\right)=56^{\circ}$

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3 years ago

Ne(g) effuses at a rate that is ______ times that of xe(g) under the same conditions.

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<span>The rate of effusion of a gas is inversely proportional to the square root of the molecular weight of the species. Now there will be differences among isotopomers but neglecting these and taking the avg mol wt of N2 = 28 and Xe = 132; Rate(N2)/Rate(Xe) = sqrt (132/28) = 2.17</span>

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3 years ago

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3 years ago

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