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3 years ago

What is the mass (in grams) of 8.11×1026 CO2 molecules?

Mathematics

1 answer:

Kobotan [32]3 years ago

6 0

Answer:

It is 59,275.74 grams

Step-by-step explanation:

• Molecular mass of carbon dioxide

$= (12 \times 1) + (16 \times 2) \\ = 12 + 32 \\ = 44 \: g$

[ molar masses: C = 12 g, O = 16 g ]

• From avogadro's number, it says 1 mole contains 6.02 × 10^23 molecules or atoms.

But 1 mole is equivalent to its molar mass in terms of weight, therefore;

${ \sf{6.02 \times {10}^{23} \: molecules = 44 \: g }}\\ { \sf{8.11 \times {10}^{26} \: molecules =( \frac{8.11 \times {10}^{26} }{6.02 \times {10}^{23} } \times 44) \: g}} \\ \\ = { \boxed{ \boxed{ \bf{59275.7 \: \: g}}}}$

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Answer:

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3 years ago

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Find the exact value of cos theta, given that sin thetaequalsStartFraction 15 Over 17 EndFraction and theta is in quadrant II.

vova2212 [387]

Answer:

$cos \theta = -\frac{8}{17}$

Step-by-step explanation:

For this case we know that:

$sin \theta = \frac{15}{17}$

And we want to find the value for $cos \theta$ , so then we can use the following basic identity:

$cos^2 \theta + sin^2 \theta =1$

And if we solve for $cos \theta$ we got:

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And if we replace the value given we got:

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For our case we know that the angle is on the II quadrant, and on this quadrant we know that the sine is positive but the cosine is negative so then the correct answer for this case would be:

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3 years ago

The ratio of two numbers is 4:5 if the larger number is 300 what is the smaller number

yaroslaw [1]

Answer:

The smaller number is 240.

Step-by-step explanation:

The question says that the larger number is 300, so 5 has to be 300.

4:5

?:300

You can use cross-multiplication or find the scale factor to find the answer.

Cross-multiplication:

4 × 300 ÷ 5 = 240

Finding the scale factor:

300 ÷ 5 = 60

4 × 60 = 240

The smaller number is 240.

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2 years ago

Sheldon earned $245.00 on Monday He is paid $35.00 per hour Whe

fredd [130]

Answer: 3:45PM

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3 years ago

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Find dy/dx if y =x^3+5x+2/x²-1

stiks02 [169]

Differentiate using the Quotient Rule –

$\qquad$ $\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\$

According to the given question, we have –

f(x) = x^3+5x+2
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Let's solve it!

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2) \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5) - ( x^3+5x+2) 2x}{(x^2-1)^2 }\\$

$\qquad$ $\pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\$

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\$

$\qquad$ $\pink{\therefore \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}} = \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\$

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2 years ago