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3 years ago

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An aluminum plate is 78 cm wide and 35 cm long. The mass of the plate is 44,226 g. Determine its thickness. [The density of lead

is 2.70g/cm^3]

1 answer:

Artemon [7]3 years ago

6 0

Answer:

i try to solve it for you, and i aplode a similer question for you

Explanation:

You might be interested in

Can an ether form hydrogen bonds in the same way as does an ester

belka [17]

Yes, an OH group from ethanol can form a hydrogen bond to the ether O atom in the same way as it can do so with the single-bonded O atom in the ester.

The O atom in the carbonyl group of the ester can also form H-bonds with ethanol.

5 0

3 years ago

Caffeine has the molecular formula, C8H10N4O2. What is the percent composition of caffeine?

Debora [2.8K]

Answer:

b) 49.48% C, 5.19% H, 28.85% N, and 16.48% O

Explanation:

we find the mass for each element in one mole by multiplying the number of atoms in one molecule with the atomic mass

mC=8Ac=8*12=96g

mH=10AH=10*1=10g

mN=4AN=4*14=56g

mO=2AO=2*16=32g

by adding the masses together we find the molar mass of the molecule

M=mC+nH+mN+mO=96+10+56+32=194g/mole

we apply the rule of threes to find the percentage of each element

194g..96gC..10gH...56gN....32gO

100g....a...........b...........c.............d

a=(100*96)/194=49.48%C

b=(100*10)/194=5.19%H

c=(100*56)/194=28.85%N

d=(100*32)/194=16.48%O

3 0

3 years ago

A sample originally contained 1.28 g of a radioisotope. It now contains 1.12 g of its daughter isotope.

dexar [7]

The answer is 3.

<span>The relation between number of half-lives (n) and decimal amount remaining (x) can be expressed as:

</span> $(1/2) ^{n} =x$

We need to calculate n, but we need x to do that. To calculate what p<span>ercentage of a radioactive species would be found as daughter material, we must calculate what amount remained:
1.28 -</span> 1.12 = 0.16

If 1.28 is 100%, how much percent is 0.16:
1.28 : 100% = 0.16 : x
x = 12.5%
Presented as decimal amount:
x = 0.125

Now, let's implement this in the equation:

<span> $(1/2) ^{n} =0.125$
</span>
Because of the exponent, we will log both sides of the equation:
$n * log(1/2) = log(0.125)$
$n = \frac{log(0.125)}{log(1/2)}$
<span> $n = \frac{log(0.125)}{log(0.5)}$
</span> $n= \frac{-0.903}{-0.301}$
$n = 3$

Therefore, 3 half-lives have passed <span> since the sample originally formed.</span>

4 0

3 years ago

Read 2 more answers

Which family (metals, nonmetals, or inert gases) generally has the valence electrons which are the easiest to remove from the at

zubka84 [21]

<span>Which family (metals, nonmetals, or inert gases) generally has the valence electrons which are the easiest to remove from the atom?
</span>
The Correct Answer Is: Metals

Hope I Helped :D

Brainliest?

-Nullgaming650

4 0

3 years ago

Read 2 more answers

You are asked to prepare 500 mL 0.300 M500 mL 0.300 M acetate buffer at pH 4.904.90 using only pure acetic acid ( MW=60.05 g/mol

defon

Answer:

You will need 9,0 g of acetic acid

Explanation:

The equilibrium acetate-acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka = 4,76

Using Henderson-Hasselbalch you will obtain:

pH = pka + log₁₀ $\frac{[A^{-}]}{[HA]}$

Where HA is acetic acid and A⁻ is acetate ion

4,90 = 4,76 + log₁₀ $\frac{[A^{-}]}{[HA]}$

1,38 = $\frac{[A^{-}]}{[HA]}$ <em>(1)</em>

As acetate concentration is 0,300M:

0,300M = [HA] + [A⁻] <em>(2)</em>

Replacing (2) in (1):

[HA] = 0,126 M

And:

[A⁻] = 0,174 M

As you need to produce 500 mL:

0,5 L × 0,126 M = 0,063 moles of acetic acid

0,5 L × 0,174 M = 0,087 moles of acetate

To produce moles of acetate from acetic acid:

CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O

Thus, moles of acetate are equivalents to moles of NaOH and all acetates comes from acetic acid, thus:

0,087 moles of acetate + 0,063 moles of acetic acid ≡ 0,15 moles of acetic acid × $\frac{60,05 g}{1mol}$ = <em>9,0 g of acetic acid</em>

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I hope it helps!

5 0

3 years ago