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3 years ago

For the skeletal chemical equation

1 answer:

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If a stone with an original velocity of 0 is falling from a ledge and takes 8 seconds to hit the ground what is the final veloci

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Also 0 as it hits the ground and stops

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3 years ago

The pH of a solution is 2.0. Which statement is correct?

Natalka [10]

Answer:

The relationship of pH and concentration of H+ ion is pH = -lg[H+]. So the concentration of H+ is 10^(-2). And [OH-][H+]=10^(-14). pOH + pH = 14. So the right answer is A.

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3 years ago

Read 2 more answers

A gas is confined in a 0.3m diameter cylinder by a piston, on which rests a weight. The mass of the piston is 85 kg. The local a

iris [78.8K]

Explanation:

Force applied on the gas will be as follows.

$F_{gas} = F_{atm} + (m + M) g$

As, F = pressure × area. Hence, calculate the forces as follows.

$F_{gas}$ = pressure × area

= $1.4 \times 10^{5} Pa \times \pi \times (\frac{0.3}{2})^{2}$

= $1.979 \times 10^{4}$ N

$F_{atm}$ = pressure × area

= $1.0133 \times 10^{5} \times \pi \times (\frac{0.3}{2})^{2}$

= $1.432 \times 10^{4}$ N

$F_{gas}$ - $F_{atm}$ = $5.47 \times 10^{3}$ N

Substituting the calculated values into the above formula as follows.

$F_{gas} = F_{atm} + (m + M) g$

$F_{gas} - F_{atm}$ = (m + M) g

$5.47 \times 10^{3}$ N = $(m + 85) \times 9.8$

$5.47 \times 10^{3}$ N = $9.8m + 833$

m = 472.76 kg

Thus, we can conclude that the mass is 472.76 kg.

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4 years ago

An element has an average atomic mass of 1.008 amu. It consists of two isotopes , one having a mass of 1.007 amu, and one having

dimulka [17.4K]

Answer:

The most abundant isotope is 1.007 amu.

Explanation:

Given data:

Average atomic mass = 1.008 amu

Mass of first isotope = 1.007 amu

Mass of 2nd isotope = 2.014 amu

Most abundant isotope = ?

Solution:

First of all we will set the fraction for both isotopes

X for the isotopes having mass 2.014 amu

1-x for isotopes having mass 1.007 amu

The average atomic mass is 1.008 amu

we will use the following equation,

2.014x + 1.007 (1-x) = 1.008

2.014x + 1.007 - 1.007 x = 1.008

2.014x - 1.007x = 1.008 - 1.007

1.007 x = 0.001

x= 0.001/ 1.007

x= 0.0009

0.0009 × 100 = 0.09 %

0.09 % is abundance of isotope having mass 2.014 amu because we solve the fraction x.

now we will calculate the abundance of second isotope.

(1-x)

1-0.0009 = 0.9991

0.9991 × 100= 99.91%

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3 years ago

At 14,000 ft elevation the air pressure drops to 0.59 atm. Assume you take a 1L sample of air at this altitude and compare it to

Ray Of Light [21]

Answer:

There are 0.1125 g of O₂ less in 1 L of air at 14,000 ft than in 1 L of air at sea level.

Explanation:

To solve this problem we use the ideal gas law:

PV=nRT

Where P is pressure (in atm), V is volume (in L), n is the number of moles, T is temperature (in K), and R is a constant (0.082 atm·L·mol⁻¹·K⁻¹)

Now we calculate the number of moles of air in 1 L at sea level (this means with P=1atm):

1 atm * 1 L = n₁ * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 K

n₁=0.04092 moles

Now we calculate n₂, the number of moles of air in L at an 14,000 ft elevation, this means with P = 0.59 atm:

0.59 atm * 1 L = n₂ * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 K

n₂=0.02414 moles

In order to calculate the difference in O₂, we substract n₂ from n₁:

0.04092 mol - 0.02414 mol = 0.01678 mol

Keep in mind that these 0.01678 moles are of air, which means that we have to look up in literature the content of O₂ in air (20.95%), and then use the molecular weight to calculate the grams of O₂ in 20.95% of 0.01678 moles:

$0.01678mol*\frac{20.95}{100} *32\frac{g}{mol} =0.1125 gO_{2}$

4 0

4 years ago