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Slav-nsk [51]
3 years ago
14

The molecular weight of NaCl is 58.44 grams/mole. If you had a 1.0 molar solution (1.0 M), you would have to put 58.44 g of salt

in 1.0 liter of solution. 100 mL of this solution would have grams of NaCl.
Chemistry
1 answer:
Andre45 [30]3 years ago
7 0

Answer:

5,844 grams of NaCl

Explanation:

Knowing the molecular weight 58,44 g/mole and saying 1 molar solution is 58,44 of NaCl in 1 liter of solution. 100 mL means 10% of the whole solution then we are going to have 10% of NaCl

58,44 x 0,1 = 5,844 grams of NaCl

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Fill in the blanks (Balancing chemical reactions)<br>BRAINLIEST
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The 2 in front of Na in 2Na + Cl_2 \rightarrow  2NaCl is <u>Coefficient.</u>

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3 years ago
Classical bonding in NaH2PO4
lesantik [10]

NaH_2PO_4, a crystal structure with a short symmetrical hydrogen bond.

<h3>What is Classical bonding?</h3>

Classical models of the chemical bond. By classical, we mean models that do not take into account the quantum behaviour of small particles, notably the electron. These models generally assume that electrons and ions behave as point charges which attract and repel according to the laws of electrostatics.

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brainly.com/question/10777799

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7 0
2 years ago
What is the wavelength of light with 2.89 x 10-19 J of energy? (The speed of
Alika [10]

Let's see

\\ \rm\rightarrowtail E=hv

\\ \rm\rightarrowtail E=\dfrac{hc}{\lambda}

  • lambda is wavelength
  • h is Planck constant
  • c is velocity of sound in air
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\\ \rm\rightarrowtail \lambda=\dfrac{hc}{E}

\\ \rm\rightarrowtail \lambda=\dfrac{6.626\times 10^{-34}\times 3\times 10^8}{2.89\times 10^{-19}}

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\\ \rm\rightarrowtail \lambda=68.8\mu m

3 0
2 years ago
You are requested to reduce the size of 50 ton/hr of a given solid. The size of the feed is such 80% passes a 4-in (76.2 mm) scr
defon

Answer:

1) The power needed to process 50 ton/hr is 135.4 HP.

2) The void fraction of the bed is 0.37.

Explanation:

1) For this type of milling operations, we can estimate the power needed for an operation according to the work index (Ei), the passing size of the circuit feed (F80) and the passing size of the product (P80).

We assume the units of Ei are kWh/t.

The equation that relates this parameters and the power is (size of particles in μm):

W=Ei*(\frac{10}{\sqrt{P80}} -\frac{10}{\sqrt{F80}} )\\\\W=9.45*(\frac{10}{\sqrt{3175\mu m}} -\frac{10}{\sqrt{76200mm}} )\\\\\\W=9.45*(0.1774+0.0362)=2.019 kWh/t

The power needed to process 50 ton/hor is

P=2.0194\frac{kWh}{Ton}*\frac{50Ton}{h}*\frac{1.341HP}{1kW}=   135.4 \, HP

2) The density of the packed bed can be expressed as

\rho=f_v*\rho_v+f_s*\rho_s

being f the fraction and ρ the density of every fraction. We know that the density of the void is 0 (ρv=0) and that fv=1-fs (the sum  of the fractions ois equal to the total space).

Then we can rearrange

\rho=f_v*\rho_v+f_s*\rho_s\\\\\rho=f_v*0+(1-f_v)*\rho_s\\\\\rho/\rho_s=1-f_v\\\\f_v=1-\rho/\rho_s=1-990/1570=1-0.63=0.37

The void fraction of the bed is 0.37.

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