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Alenkinab [10]
1 year ago
12

You are designing an office building. In order to reduce noise from high-frequency sounds between offices above and below each o

ther, would you choose carpet on concrete or wood floors? Explain your answer.
Chemistry
2 answers:
Monica [59]1 year ago
8 0

Answer:

concrete has more mass and it will block more noise than than the wood plus the wood is thiner the thicker the substance is the better u won't hear noise

Explanation:

miskamm [114]1 year ago
7 0
Carpet for sure. Have you ever heard how loud a dog walking on wood compared to carpet is.
You might be interested in
Consider the reaction between nis2 and o2: 2nis2(s)+5o2(g)→2nio(s)+4so2(g) when 11.2 g of nis2 are allowed to react with 5.43 g
stellarik [79]
M(NiS₂) = 11.2 g.
n(NiS₂) = m(NiS₂) ÷ M(NiS₂).
n(NiS₂) = 11.2 g ÷ 122.8 g/mol.
n(NiS₂) = 0.091 mol.
m(O₂) = 5.43 g.
n(O₂) = 5.43 g ÷ 32 g/mol.
n(O₂) = 0.17 mol; limiting reactant.
From chemical reaction: n(NiS₂) : n(O₂) = 2 : 5.
0.091 mol : n(O₂) = 2 : 5.
n(O₂) = 0.2275 mol, not enough.
n(NiO) = 4.89 g .
n(O₂) : n(NiS) = 5 : 2.
n(NiS) = 0.068 mol.
m(NiS) = 0.068 mol · 74.7 g/mol = 5.08 g.
percent yield = 4.89 g / 5.08 g · 100% = 96.2%.


6 0
2 years ago
Why does iron not fit the pattern in column 7
son4ous [18]
Iron doesn't fit because it doesn't have enough atoms or protons in its nucleus so there for it belongs in column 2. <span />
3 0
2 years ago
Iodine-131 has a half-life of 8.10 days. In how many days will 50 grams of Iodine-131 decay to one-eighth of its original amount
shtirl [24]
What you need to do is find 1/8 of 50
you can just divide 50 by 8 to get 6.25
so now you have to find how many days it will take till there are 6.25 grams of iodine left
every 8.1 days its mass is split in half 
so start splitting it in half and every time you do, you add 8.1 days
50/2 =25                                               8.1
25/2 =12.5                                        +  8.1
12.5/2= 6.25                                      +8.1
now you have reached 1/8 of the original amount of Iodine-131
so to find how long it took just add 8.1+8.1+8.1
(this is the same as 8.1x3)
which equals 24.3
it will take 24.3 days for Iodine 131 to decay to 1/8  of its original mass.

(good luck on the regent if thats what your studying for :)

5 0
2 years ago
Read 2 more answers
One of the most desirable of the old British sports cars was the beautiful Triumph Vitesse (1963-1971). Pictured below is the Am
vodka [1.7K]

Answer:

V_{CO_2}=1.55x10^{5}LCO_2=155m^3CO_2

Explanation:

Hello,

In this case, given the reaction:

2 C_8H_{18}(l) + 25 O_2(g)\rightarrow 16 CO_2(g) + 18 H_2O(g)

The total consumed gallons are computed by considering 686 miles were driven and the consumption is 21.2 miles per gallon, thus:

V_{C_8H_{18}}=686miles*\frac{1gal}{21.2miles} =32.4gal

Hence, with the given density, one could compute the consumed grams and consequently moles of gasoline as well as moles that were consumed:

n_{C_8H_{18}}=32.4gal*\frac{3785.41cm^3}{1gal} *\frac{0.805g}{1cm^3} *\frac{1mol}{114g}=864.95mol C_8H_{18}

Next, since gasoline (molar mass = 114 is in a 2:16 molar relationship with the yielded carbon dioxide, we compute its produced moles as shown below:

n_{CO_2}=864.95mol C_8H_{18}*\frac{16molCO_2}{2molC_8H_{18}} =6919.6molCO_2

Finally, we could assume the given STP conditions to compute the volume of carbon dioxide, as no more information regarding the space wherein the carbon dioxide is available:

V_{CO_2}=\frac{n_{CO_2}RT}{P} =\frac{6919.6mol*0.082\frac{atm*L}{mol*K}*(0+273)K}{1atm} \\\\V_{CO_2}=1.55x10^{5}LCO_2=155m^3CO_2

Best regards.

5 0
2 years ago
A new potential heart medicine, code-named X-281, is being tested by a pharmaceutical company, Pharma-pill. As a research techni
torisob [31]

Answer:

a. pka = 3,73.

b. pkb = 10,27.

Explanation:

a. Supposing the chemical formula of X-281 is HX, the dissociation in water is:

HX + H₂O ⇄ H₃O⁺ + X⁻

Where ka is defined as:

ka = \frac{[H_3O^+][X^-]}{[HX]}

In equilibrium, molar concentrations are:

[HX] = 0,089M - x

[H₃O⁺] = x

[X⁻] = x

pH is defined as -log[H₃O⁺]], thus, [H₃O⁺] is:

[H_3O^+]} = 10^{-2,40}

[H₃O⁺] = <em>0,004M</em>

Thus:

[X⁻] = 0,004M

And:

[HX] = 0,089M - 0,004M = <em>0,085M</em>

ka = \frac{[0,004][0,004]}{[0,085]}

ka = 1,88x10⁻⁴

And <em>pka = 3,73</em>

b. As pka + pkb = 14,00

pkb = 14,00 - 3,73

<em>pkb = 10,27</em>

I hope it helps!

4 0
2 years ago
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