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1 year ago

USATEST PREP QUESTION HELP ! Really appreciate it !

Chemistry

2 answers:

dezoksy [38]1 year ago

8 0

Answer:

The answer is A.

Novosadov [1.4K]1 year ago

5 0

Answer is A Access pictures of the area taken by satelites.

Explanation: Satelites are the only thing out of these four answers that does not requir power supply from the town. Hope it helped!

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If charges flow very slowly through a metal, why does it not require several hours for a light to come on when you throw a switc

Harlamova29_29 [7]

T<span>he charges themselves may not move fast, but the force upon them does. The electric field set up by the battery or generator propagates through the wires at the speed of light.

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
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6 0

2 years ago

A bar of gold is 5.0mm thick, 10.0cm long and 2.0cm wide. It has a mass of exactly 193.0g. What is the desity of gold?

Tanzania [10]

<h3>Answer:</h3>

19.3 g/cm³

<h3>Explanation:</h3>

Density of a substance refers to the mass of the substance per unit volume.

Therefore, Density = Mass ÷ Volume

In this case, we are given;

Mass of the gold bar = 193.0 g

Dimensions of the Gold bar = 5.00 mm by 10.0 cm by 2.0 cm

We are required to get the density of the gold bar

Step 1: Volume of the gold bar

Volume is given by, Length × width × height

Volume = 0.50 cm × 10.0 cm × 2.0 cm

= 10 cm³

Step 2: Density of the gold bar

Density = Mass ÷ volume

Density of the gold bar = 193.0 g ÷ 10 cm³

= 19.3 g/cm³

Thus, the density of the gold bar is 19.3 g/cm³

3 0

1 year ago

Butyl butyrate is an ester that is a naturally occurring oil used in the flavor industry for its fruity scent. If the steam dist

Snezhnost [94]

Answer:

The correct answer is 62.5 %.

Explanation:

Based on the given information, the partial pressure of butyl butyrate is 50 mmHg and the partial pressure of water is 710 mmHg.

Hence, the total pressure is 710+50 = 760 mmHg

According to Dalton's law of partial pressure,

Partial pressure = mole fraction * total pressure

Mole fraction of water is,

Partial pressure of water/Total pressure = 710/760 = 0.93

Similarly, the mole fraction of butyl butyrate is,

Partial pressure of butyl-butyrate/Total pressure = 50/760 = 0.07

Therefore, mole% of water is 0.93 * 100 = 93 %

For calculating mass%,

Mass of H2O = 0.93 * 18 = 16.8 grams (The molecular mass of water is 18 grams per mole)

The molecular mass of butyl-butyrate is 144 gram per mole

The mass of butyl-butyrate = 144 * 0.07 = 10.08 grams

The mass percent of water will be,

Mass % of water/Total mass % * 100 = 16.8 / 10.08 + 16.8 * 100 = 62.5%.

8 0

1 year ago

Use the thermodynamic data at 298 k below to determine the ksp for barium carbonate, baco3 at this temperature. substance: ba2+(

jeka57 [31]

Answer : the correct answer for ksp = 1.59 * 10⁻⁹

Following are the steps to calculate the ksp of reaction

BaCO₃ →Ba ²⁺ + CO₃²⁻ :

Step 1 : To find ΔG° of reaction :

ΔG° of reaction can be calculates by taking difference between ΔG° of products and reactants as :

ΔG° reaction =Sum of ΔG° ( products ) - Sum of Δ G° ( reactants ) .

Given : ΔG° for Ba²⁺ ( product )= -560.7 $\frac{KJ}{Mol}$

ΔG° for CO₃²⁻ (product ) =- 528.1 $\frac{KJ}{Mol}$

ΔG° BaCO₃ ( reactant) = –1139 $\frac{KJ}{Mol}$

Plugging value in formula :

ΔG° for reaction = ( ΔG° of Ba ²⁺ + ΔG° of CO₃²⁻ ) - (ΔG° of BaCO₃ )

⁻ = ( -560.7 $\frac{KJ}{Mol}$ + 528.1 $\frac{KJ}{Mol}$ ) - ( -1139 $\frac{KJ}{Mol}$ )

= ( -1088.8 $\frac{KJ}{Mol}$ ) - (-1139 $\frac{KJ}{Mol}$ )

= - 1088.8 $\frac{KJ}{Mol}$ + 1139 $\frac{KJ}{Mol}$

ΔG° of reaction = 50.2 $\frac{KJ}{Mol}$

Step 2: To calculate ksp from ΔG° of reaction .

The relation between Ksp and ΔG° is given as :

ΔG° = -RT ln ksp

Where ΔG° = Gibb's Free energy R = gas constant T = Temperature

Ksp = Solubility constant product .

Given : ΔG° of reaction = 50.2 $\frac{KJ}{Mol}$

T = 298 K R = 8.314 $\frac{J}{Mol * K}$

Plugging values in formula

$50.2 \frac{KJ}{mol} = - 8.314 \frac{J}{mol * K} * 298 K * ln ksp$

$50.2 \frac{KJ}{mol} = - 2477.572 \frac{J}{mol} * ln K$

((Converting 2477 $\frac{J}{mol} to \frac{KJ}{mol}$

Since , 1 KJ = 1000 J So , $2477 \frac{J}{mol} * \frac{1 KJ}{1000J} = 2.477 \frac{KJ}{mol}$ ))

Dividing both side by $- 2.477 \frac{KJ}{mol}$

$\frac{50.2\frac{KJ}{mol}}{-2.477 \frac{KJ}{mol}} = \frac{-2.477 \frac{KJ}{mol}}{-2.477 \frac{KJ}{mol}} * ln ksp$

ln ksp = $ln ksp = -20.27 \frac{KJ}{mol}$

Removing ln :

ksp = 1. 59 * 10⁻⁹

8 0

2 years ago

For the following dehydrohalogenation (E2) reaction, draw the Zaitsev product(s) resulting from elimination involving C3–C4 (i.e

Bond [772]

Answer:

2-methyl-butene

Explanation:

For the E2 mechanism, we have an <u>anti-elimination</u>. The Br leaves the molecule and the base removes the hydrogen in the anti position to form the double that's why only one structure is produced. (See figure 1)

Since we have 2 hydrogens on the right carbon, we cannot indicate a <u>specific stereoisomer</u>, in other words, it is not possible to assign a <u>Z or E</u> configuration for this alkene.

8 0

2 years ago