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3 years ago

15

some constellations, such ad Ursa Minor, are visible in the sky year-round; other contellations appear for only part of the year

. explain why this happens.

1 answer:

worty [1.4K]3 years ago

5 0

While most constellations are only visible to us in different seasons, some are always there 24/7/365 because they are positioned close to the Polar Axis, or the Polaris.

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What is the total gas pressure in a sealed flask that contains oxygen at a partial pressure of 0.35 atm and water vapor at a par

Fofino [41]

According to Dalton's Law, in a mixture of non-reacting gasses, thetotal pressure<span> exerted is the sum of the </span>partial pressures<span> of the component gasses. In more complicated circumstances, equilibrium states come into effect, but fortunately for us, </span>oxygen<span> is non-reactive with </span>water vapor<span>.</span>

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3 years ago

How many grams of iodine are needed to prepare 28.6 grams of ICl

Maslowich

The answer will be 17.9 grams

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3 years ago

It has been suggested that the surface melting of ice plays a role in enabling speed skaters to achieve peak performance. Carry

vesna_86 [32]

Explanation:

Relation between pressure, latent heat of fusion, and change in volume is as follows.

$\frac{dP}{dT} = \frac{L}{T \times \Delta V}$

Also, $\frac{L}{T} = \Delta S^{fusion}_{m}$

where, $\Delta V^{fusion}_{m}$ is the difference in specific volumes.

Hence, $\frac{dP}{dT} = \frac{\Delta S^{fusion}_{m}}{\Delta V^{fusion}_{m}}$

As, $\Delta S^{fusion}_{m} = \frac{L}{T} = \frac{6010}{273.15}$ = 22.0 J/mol K

And, $\Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}$ ...... (1)

where, $d_{H_{2}O}$ = density of water

$d_{ice}$ = density of ice

M = molar mass of water = $18.02 \times 10^{-3} kg$

Therefore, using formula in equation (1) we will calculate the volume of fusion as follows.

$\Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}$

= $\frac{18.02 \times 10^{-3}}{997} - \frac{18.02 \times 10^{-3}}{920}$

= $-1.51 \times 10^{-6}$

Therefore, calculate the required pressure as follows.

$\frac{dP}{dT} = \frac{22}{-1.51 \times 10^{-6}}$

= $1.45 \times 10^{7} Pa/K$

or, = 145 bar/K

Hence, for change of 1 degree pressure the decrease is 145 bar and for 4.7 degree change dP = $145 \times 4.7 bar$

= 681.5 bar

Thus, we can conclude that pressure should be increased by 681.5 bar to cause 4.7 degree change in melting point.

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3 years ago

ASAP!!!!!!!!!!!!

dlinn [17]

<span>a. Tall prarie grass burns after being struck by lightning.</span>

6 0

3 years ago

Given that kb for c6h5nh2 is 1.7 × 10-9 at 25 °c, what is the value of ka for c6h5nh3 at 25 °c?

lina2011 [118]

Answer: $k_a$ for $C_6H_5NH_3^+$ at 25°C is $0.588\times 10^{-4}$

Explanation: We are given $k_b$ of $C_6H_5NH_2$ at 25°C which is $1.7\times 10^{-9}$

To calculate the $k_a$ of $C_6H_5NH_3^+$ , we use the formula:

$k_w=k_a\times k_b$

$k_w\text{ at }25^o=1\times 10^{-14}$

Putting values in above equation, we get:

$1\times 10^{-14}=k_a\times (1.7\times 10^{-9})\\k_a=0.588\times 10^{-5}$

7 0

3 years ago