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Alla [95]
3 years ago
14

A 76.00-pound flask of mercury costs $151.50. The density of mercury is 13.534 g/cm3.

Chemistry
2 answers:
Agata [3.3K]3 years ago
8 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Please see below solution:

1 lb Hg x (151.50/76 lb) = cost of 1 lb. 
<span>cost 1 lb x (1g/453.6 g) = cost of 1 g.</span>
gulaghasi [49]3 years ago
6 0

Answer:

cost of 1 cm^{3} mercury is \$ 0.059477

cost of 1 in^{3} mercury is \$ 0.97466

cost of 1 pound mercury is \$ 1.99

cost of 1 g mercury is \$ 0.00439

Explanation:

We know, density = \frac{mass}{volume}

1 pound =453.6 g

So, mass of 76.00 pound mercury = (76.00\times 453.6)g=34473.6g

So, volume of 34473.6 g mercury = \frac{34473.6g}{13.534g/cm^{3}}= 2547.2 cm^{3}

So, cost of 2547.2 cm^{3} mercury is \$ 151.50

Hence cost of 1 cm^{3} mercury is \frac{\$ 151.50}{2547.2}=\$ 0.059477

1 cm^{3} = 0.061024 in^{3}

So, cost of 0.061024 in^{3} mercury is \$ 0.059477

Hence, cost of 1 in^{3} mercury is \frac{\$ 0.059477}{0.061024}=\$ 0.97466

Cost of 76.00 pound mercury is \$ 151.50

So, cost of 1 pound mercury is \frac{\$ 151.50}{76.00}=\$ 1.99

Cost of 34473.6 g of mercury is  \$ 151.50

So, cost of 1 g mercury is \frac{\$ 151.50}{34473.6}=\$ 0.00439

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I have a balloon that can hold 125,000 mL of air. If I blow up this balloon with 3 moles of oxygen gas at a
matrenka [14]

Answer:

234.35 °C

Explanation:

Given data:

Volume of balloon = 125000 mL

Moles of oxygen = 3 mol

Pressure = 1 atm

Temperature = ?

Solution:

Formula:

PV = nRT

P = Pressure

V = volume

n = number of moles

R = ideal gas constant

T = temperature

Volume of balloon = 125000 mL × 1 L /1000 mL

Volume of balloon = 125 L

Now we will put the values:

Ideal gas constant = R = 0.0821 atm.L/mol.K

PV = nRT

T = PV/nR

T = 1 atm × 125 L/  0.0821 atm.L/mol.K × 3 mol

T= 125  /0.2463 /K

T = 507.5 K

K to °C

507.5 K - 273.15 = 234.35 °C

4 0
3 years ago
If the pKa value for a given acid is low, what does that say about the acid (is it weak or strong)?
Dmitry [639]

Answer:

If the pKa of the acid is low (negative), then the acid is strong.

Explanation:

Ka, <em>the acid ionization constant,  </em>measures the strength of an acid in a solution. Stronger acids have higher Ka values.

We defined: pKa = -log[Ka]

This function is a decreasing function, meaning that pKa will be getting smaller and smaller,  while increasing Ka (high values of Ka will have negative pKa values). Therefore, stronger acids (high values of Ka), will have low (negative) pKa values.

3 0
4 years ago
Refer to the first three rows of the periodic table, what element has properties most similar to carbon?
forsale [732]
Silicone is most similar to carbon.
 hope this helps; and good luck.
8 0
3 years ago
Which remains the same as the distance of an object from Earth changes?
zvonat [6]

Answer:

its mass which is C

Explanation:

3 0
4 years ago
The solubility of o2 at 20c is 1.38 x10^-3. the partial presure of o2 in the air at sea level is 0.27 atm. using henery;s law, c
netineya [11]

<u>Answer:</u> The solubility of oxygen at 682 torr is 4.58\times 10^{-3}M

<u>Explanation:</u>

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_{A}=K_H\times p_{A}

Or,

\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}

where,

C_1\text{ and }p_1 are the initial concentration and partial pressure of oxygen gas

C_2\text{ and }p_2 are the final concentration and partial pressure of oxygen gas

We are given:

Conversion factor used:  1 atm = 760 torr

C_1=1.38\times 10^{-3}M\\p_1=0.27atm\\C_2=?\\p_2=682torr=0.897atm

Putting values in above equation, we get:

\frac{1.38\times 10^{-3}}{C_2}=\frac{0.27atm}{0.897atm}\\\\C_2=\frac{1.38\times 10^{-3}\times 0.897atm}{0.27atm}=4.58\times 10^{-3}M

Hence, the solubility of oxygen gas at 628 torr is 4.58\times 10^{-3}M

4 0
3 years ago
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