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Morgarella [4.7K]
3 years ago
13

The partial pressure of carbon dioxide in the atmosphere is 0.239 torr. Calculate the partial pressure in mm Hg and cant Round e

ach of your answers to 3 alg X 5 ?
Chemistry
1 answer:
victus00 [196]3 years ago
6 0

Explanation:

It is known that 1 torr equals 1 mm Hg.

Therefore, 0.239 torr will be converted into mm Hg as follows.

                     1 torr = 1 mm Hg

So,          0.239 torr = 0.239 torr \times \frac{1 mm Hg}{1 torr}

                                 = 0.239 mm Hg

It means that 0.239 torr is equal to 0.239 mm Hg.

Therefore, we can conclude that partial pressure of given carbon dioxide in mm Hg is 0.239 mm Hg.

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4 years ago
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Density is d= m/vol. If a material has a mass of 65.5 g and a volume of 32.5 ml, it has a density of
ryzh [129]

Using the given formula, the density of the material is 2.015 g/mL

<h3>Calculating Density </h3>

From the question, we are to determine the density of the material

From the given formula

Density = Mass / Volume

And from the given information,

Mass = 65.5 g

and volume = 32.5 mL

Putting the parameters into the equation,

Density = 65.5/32.5

Density = 2.015 g/mL

Hence, the density of the material is 2.015 g/mL.

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8 0
2 years ago
Given the chemical equation: 2 Pb + O2 → 2 PbO, if 51.8 grams of Pb are formed in this reaction, then 8.00 grams of O2 must have
Nutka1998 [239]

Answer:

If 51.8 of Pb is reacting, it will require 4.00 g of O2

If 51.8 g of PbO is formed, it will require 3.47 g of O2.

Explanation:

Equation of the reaction:

2 Pb + O2 → 2 PbO

From the equation of reaction, 2 moles of lead metal, Pb, reacts with 1 mole of oxygen gas, O2, to produce 2 moles of lead (ii) oxide, PbO

Molar mass of Pb = 207 g

Molar mass of O2 = 32 g

Molar mass of PbO = 207 + 32 = 239 g

Therefore 2 × 207 g of Pb reacts with 32 g of O2 to produce 2 × 239 g of PbO

= 414 g of Pb reacts with 32 g of O2 to produce 478 g of PbO

Therefore, formation of 51.8 g of PbO will require (32/478) × 51.8 of O2 = 3.47 g of O2.

If 51.8 of Pb is reacting, it will require (32/414) × 51.8 g of O2 = 4.00 g of O2

7 0
3 years ago
What is the ph of a solution of 1.699 l of 1.25 m hcn, ka = 6.2 x 10-10, and 1.37 moles of nacn?
BlackZzzverrR [31]

The pH of a solution is 9.02.

c(HCN) = 1.25 M; concentration of the cyanide acid

n(NaCN) = 1.37 mol; amount of the salt

V = 1.699 l; volume of the solution

c(NaCN) = 1.37 mol ÷ 1.699 l

c(NaCN) = 0.806 M; concentration of the salt

Ka = 6.2 × 10⁻¹⁰; acid constant

pKa = -logKa

pKa = - log (6.2 × 10⁻¹⁰)

pKa = 9.21

Henderson–Hasselbalch equation for the buffer solution:

pH = pKa + log(cs/ck)

pH = pKa + log(cs/ck)

pH = 9.21 + log (0.806M/1.25M)

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8 0
1 year ago
If you added 45,000 calories to water that was at 25 degrees C, and the ending temperature was 35 degrees C, how much water did
user100 [1]

<u>Answer:</u>

<em>4.5 L water we have in litres (L).</em>

<em><u></u></em>

<u>Explanation:</u>

Q=m\times c \times \Delta T

where

\Delta T = Final T - Initial T

Q is the heat energy in calories

c is the specific heat capacity (for water 1.0  cal/(g℃))  

m is the mass of water

Plugging in the values  

\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times\left(35^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}\right)$\\\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}$\\\\$m=\frac{45000 \mathrm{cal}}{1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}}$\\\\$m=4500 \mathrm{g}$\\\\Density of water $=\frac{\text { mass }}{\text { volume }}$

So,

Volume of water = mass/density

\\\\=\frac{4500 \mathrm{g}}{\frac{1.09}{\mathrm{mL}}}=4500 \mathrm{mL}$$

=4.5 L (Answer)

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4 years ago
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