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2 years ago

Part 1 Designing an Investigation

Chemistry

1 answer:

Kobotan [32]2 years ago

3 0

Answer:

The experimental plan is to measure the values of the dependent variable, which is the temperature of the pizza after it is cooled in each of the heat (temperature) environments, which is the dependent variable, for a given equal period of time, which is the control

Explanation:

The given parameters are;

The temperature of the pizza = 400°F

The temperature of the freezer = 0°F

The temperature of the refrigerator = 40°F

The temperature of the countertop = 78°F

Given that the independent variable = The heat to which the hot pizza is subjected

The dependent variable = The temperature to which the pizza cools down

The experiment plan includes;

1) Place the pizza which is at 400°F in each of the different heat environment, which are, the freezer, the fridge, and the counter top, for the same period of time and record the final temperature of the pizza

2) The option that gives the lowest final temperature within the same time frame is the option that will let the pizza cool down fastest.

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What occurs when an optically active alcohol reacts with HBr to give an alkyl halide? Multiple Choice the chirality center retai

denpristay [2]

Answer:

The incomplete and varying inversion of configuration takes place at the chirality center.

Explanation:

When optically active alcohols react with HBr an SN1 reaction occurs.

In SN1 reactions an intermediate carbocation is formed in which the nucleophile can attack it on either side of the molecule. Therefore, there is a partial inversion of the center of chilarity of the molecule.

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2 years ago

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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.

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2 years ago

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The specific rotation of (R) carvone is (+) 61°. The optical rotation of a sample of a mixture of R &S carvone is measured a

shusha [124]

Answer:

See explanation

Explanation:

% optical purity = specific rotation of mixture/specific rotation of pure enantiomer * 100/1

specific rotation of mixture = 23°

specific rotation of pure enantiomer = 61°

Hence;

% optical purity = 23/61 * 100 = 38 %

More abundant enantiomer = 100% - 38 % = 62%

Hence the pure (S) carvone is (-) 62° is the more abundant enantiomer.

Enantiomeric excess = 62 - 50/50 * 100 = 24%

Hence

(R) - carvone = 38 %

(S) - carvone = 62%

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1 year ago

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Pesticide concentrations in the Rhine River between Germany and France between 1969 and 1975 averaged 0.55 mg/L of hexachloroben

sertanlavr [38]

Answer:

1.93×10⁻³ mmoles/L of C₆Cl₆; 1.58×10⁻⁴ mmoles/L of C₁₂H₈Cl₆O; 3.51×10⁻³ mmoles/L of C₆H₆Cl₆

Explanation:

We have to find out the molar mass of each pesticide to calculate the moles, and then the milimoles

C₆Cl₆ → 12. 6 + 35.45 .6 = 284.7 g/m

C₁₂H₈Cl₆O → 12 . 12 + 8.1 + 35.45 .6 + 16 = 380.7 g/m

C₆H₆Cl₆ → 12.6 + 6.1 + 35.45 .6 = 290,7 g/m

Let's convert mg to g (/1000)

0.55 mg / 1000 = 5.5×10⁻⁴ g

0.060 mg / 1000 = 6×10⁻⁵ g

1.02 mg / 1000 = 1.02×10⁻³ g

Now we can know the moles (mass / molar mass)

5.5×10⁻⁴ g / 284.7 g/m = 1.93×10⁻⁶ moles of C₆Cl₆

6×10⁻⁵ g / 380.7 g/m = 1.58×10⁻⁷ moles of C₁₂H₈Cl₆O

1.02×10⁻³ g / 290,7 g/m = 3.51×10⁻⁶ moles of C₆H₆Cl₆

Milimoles = Mol . 1000

1.93×10⁻⁶ . 1000 = 1.93×10⁻³ mmoles of C₆Cl₆

1.58×10⁻⁷ . 1000 = 1.58×10⁻⁴ mmoles of C₁₂H₈Cl₆O

3.51×10⁻⁶ . 1000 = 3.51×10⁻³ mmoles of C₆H₆Cl₆

6 0

2 years ago

Calculate the mass of 25,000 molecules of nitrogen gas. (1 mole = 6.02 x 1023 molecules)

Ainat [17]

Hey there!

Molar mass N2 = 28.01 g/mol

Therefore:

28.01 g N2 -------------- 6.02*10²² molecules N2

( mass N2 ?? ) ----------- 25,000 molecules N2

mass N2 = ( 25,000 * 28.01 ) / ( 6.02*10²³ )

mass N2 = 700250 / 6.02*10²³

mass N2 = 1.163*10⁻¹⁸ g

Hope that helps!

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2 years ago

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