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Elan Coil [88]
3 years ago
12

How much cooking liquid is needed when employing the pasta method?

Chemistry
2 answers:
Sonbull [250]3 years ago
8 0
A l l of IT? idk im 7
Crank3 years ago
8 0

Answer: A lot, but depending of how much pasta/rice you´re cooking.

Explanation: There are several users of the pasta method around the Internet that define the correct amount of water with "what seems right to you", and believe it or not this migh not be a lie at all. One of the constants that I've found on this method is the instinct. Put as much water as you wish, but your boiling vessel has to look good! A recipe that generates crave at sight.

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Carry out the following calculation, paying special attention to the significant figures (where 4/3 is exact), rounding, and uni
babunello [35]

Answer:

Value = 1.80 g/cm³ (Approx)

Explanation:

Given:

\frac{3.39 \times 10^7g}{(\frac{4}{3} )(3.1416)(1.65 \times 10^2 cm)^3}

Computation:

\frac{3.39 \times 10^7g}{(\frac{4}{3} )(3.1416)(1.65 \times 10^2 cm)^3} \\\\\frac{3.39 \times 10^7g}{(\frac{4}{3} )(3.1416)(4.492125 \times 10^6 cm^3)} \\\\ \frac{3.39 \times 10^7g}{(\frac{4}{3} )(3.1416)(4.492125 \times 10^6 cm^3)}\\\\ \frac{3.39 \times 10^7g}{18.8166132\times 10^6 cm^3} \\\\ 1.80159945g/cm^3

Value = 1.80 g/cm³ (Approx)

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3 years ago
Find the mistake<br> Cu(NO3)2(s) + CuO(s) → 2NO(g) + O3(g)
kvv77 [185]

there are 2 copper atoms on the left side of the arrow, but none on the other side. there are also 7 oxygen atoms on the left hand side, but only 5 on the right hand side of the arrow.
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What animal wins march mammal madness in 2021?
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A flask with a volume of 3.16 l contains 9.33 grams of an unknown gas at 32.0°c and 1.00 atm. What is the molar mass of the gas?
Inessa05 [86]

Answer:

73.88 g/mol

Explanation:

For this question we have to keep in mind that the unknown substance is a <u>gas</u>, therefore we can use the <u>ideal gas law</u>:

PV=nRT

In this case we will have:

P= 1 atm

V= 3.16 L

T = 32 ªC = 305.15 ºK

R= 0.082 \frac{atm*L}{mol*K}

n= ?

So, we can <u>solve for "n"</u> (moles):

1~atm*3.16~L~=~n*0.082~\frac{atm*L}{mol*K}*305.15~K

n=\frac{1~atm*3.16~L~}{0.082~\frac{atm*L}{mol*K}*305.15~K}

n=0.126~mol

Now, we have to remember that the <u>molar mass value has "g/mol"</u> units. We already have the grams (9.33 g), so we have to <u>divide</u> by the moles:

molar~mass=\frac{9.33~grams}{0.126~mol}

molar~mass=73.88\frac{grams}{mol}

7 0
3 years ago
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Answer:

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9. d

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