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3 years ago

When iodobenzene codistills with water at 760 toor, the boiling point of the mixture is 98.5. the vapor pressure of water is?

1 answer:

6 0

<span>This was one of the old regents. I didn't get this at first glance too so I searched for it and I got this site. lol. I know what to do now. To get a higher boiling point, you have to find which solution has the most moles. Since SO4, PO4, and NO3 are all polyatomic ions, they only count as one mole. </span>

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What are the examples of 8 branches of chemistry?

-Dominant- [34]

Answer:

hmm suspicious

lol I don't know bye

4 0

2 years ago

In the manufacture of paper, logs are cut into small chips, which are stirred into an alkaline solution that dissolves several o

Kitty [74]

Answer:

The estimated feed rate of logs is 14.3 logs/min.

Explanation:

The product of the process is 2000 tons/day of dry wood pulp, of 85 wt% of cellulose. That represents (2000*0.85)=1700 tons/day of cellulose.

That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.

Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.

To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.

$V=V_c+V_w\\\\M/\rho=M_c/\rho_c+Mw/\rho_w\\\\M/\rho=0.55*M/\rho_c+0.45*M/\rho_w\\\\1/\rho=0.55/\rho_c +0.45/\rho_w\\\\0.55/\rho_c=1/\rho-0.45/\rho_w\\\\0.55/\rho_c=1/(0.64*\rho_w)-0.45/\rho_w=(1/\rho_w)*(\frac{1}{0.64}-\frac{0.45}{1} )\\\\0.55/\rho_c=1.1125/\rho_w\\\\\rho_c=\frac{0.55}{1.1125}*\rho_w= 0.494*\rho_w$

The specific gravity of the wood chips is 0.494.

The average volume of a log is

$V_l=(\pi*D^{2} /4)*L=(3.1416*\frac{8^{2} \, in^{2} }{4} )*9ft*(\frac{12 in}{1ft})= 21714 in^{3}=12.57 ft^{3}$

The weight of one log is

$M=\rho*V=0.494*\rho_w*12.57 ft^{3}\\\\M=0.494*62.4\frac{lbm}{ft^{3} }*12.57ft^{3}\\\\M=387.5lbm$

To provide 3617 ton/day of wood chips, we need

$n=\frac{supply}{M_{log}}=\frac{3617 tons/day}{387.5 lbm}*\frac{2204lbm}{1ton}\\\\n=20573 logs/day=14.3 logs/min$

The feed rate of logs is 14.3 logs/min.

7 0

3 years ago

Read 2 more answers

Jen makes a venn diagram to compare active transport and passive transport

balu736 [363]

Both active and passive transport move molecules

3 0

3 years ago

Read 2 more answers

The fuel used in many disposable lighters is liquid butane, C4H10. Butane has a molecular weight of 58.1 grams in one mole. How

BARSIC [14]

<h3>Answer:</h3>

6.21 × 10²² Carbon Atoms

<h3>Solution:</h3>

Data Given:

Mass of Butane (C₄H₁₀) = 1.50 g

M.Mass of Butane = 58.1 g.mol⁻¹

Step 1: Calculate Moles of Butane as,

Moles = Mass ÷ M.Mass

Putting values,

Moles = 1.50 g ÷ 58.1 g.mol⁻¹

Moles = 0.0258 mol

Step 2: Calculate number of Butane Molecules;

As 1 mole of any substance contains 6.022 × 10²³ particles (Avogadro's Number) then the relation for Moles and Number of Butane Molecules can be written as,

Moles = Number of C₄H₁₀ Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Solving for Number of Butane molecules,

Number of C₄H₁₀ Molecules = Moles × 6.022 × 10²³ Molecules.mol⁻¹

Putting value of moles,

Number of C₄H₁₀ Molecules = 0.0258 mol × 6.022 × 10²³ Molecules.mol⁻¹

Number of C₄H₁₀ Molecules = 1.55 × 10²² C₄H₁₀ Molecules

Step 3: Calculate Number of Carbon Atoms:

As,

1 Molecule of C₄H₁₀ contains = 4 Atoms of Carbon

So,

1.55 × 10²² C₄H₁₀ Molecules will contain = X Atoms of Carbon

Solving for X,

X = (1.55 × 10²² C₄H₁₀ Molecules × 4 Atoms of Carbon) ÷ 1 Molecule of C₄H₁₀

X = 6.21 × 10²² Atoms of Carbon

5 0

3 years ago

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Para que sirven las rondas??<br> ayuda por favor

prisoha [69]

Answer:

las rondas sirven para ayudar con problemas matematicos asi como cilindros

6 0

3 years ago