How do clouds affect solar energy (radiation)?

The orbital radius of Venus is 0.72 AU. What is this distance in kilometers? (One AU is about 150 million kilometers.)

Evgen [1.6K]

<span>The answer is D) 108 million kilometers. To solve this problem, you must perform a simple unit conversion calculation. 1 AU = 150,000,000 km is the conversion factor. Take the radius of Venus, .72 AU, and multiply it by 150,000,000 km/1 AU. You flip the conversion factor so that the units of the original distance in the numerator cancel the units in the denominator of the conversion factor. completing the calculation gives you 108,000,000 km</span>

6 0

3 years ago

The radioisotope xenon-133 has a half-life of 5.2 days. How much of an 80 gram sample would be left after 20.8 days? PLEASE SHOW

yarga [219]

Answer:

6.73g

Explanation:

T½ = 5.2days

No = 80g

N = ?

T = 20.8days

We'll have to find the disintegration constant first so that we can plug it into the equation that will help us find the mass of the sample after 20.8 days

T½ = In2 / λ

T½ = half life

λ = disintegration constant

λ = In2 / T½

λ = 0.693 / 5.8

λ = 0.119

In(N / No) = -λt

N = final mass of the radioactive sample

No = initial mass of the sample

λ = disintegration constant

t = time for the radioactive decay

In(N/No) = -λt

N / No = e^-λt

N = No(e^-λt)

N = 80 × e^-(0.119 × 20.8)

N = 80 × e^-2.4752

N = 80 × 0.0841

N = 6.728g

The mass of the sample after 20.8 days is approximately 6.73g

6 0

3 years ago

Which surface ocean current has the warmest water.

SOVA2 [1]

Which surface ocean current has the warmest water.
The answer would be B

5 0

4 years ago

Read 2 more answers

White wash suspension, a saturated solution of ca(oh)2, has a ph of 12.37. what is ksp for ca(oh)2?

SSSSS [86.1K]

Find the $pOH$ from $pH$

$pH + pOH =14\\pOH= 14-pH = 14-12.37 = 1.63$

The $KSP$ value of $ca(oh)^{2}$ is

$ca(oh)^{2} -- > ca^{2} + (aq) +2OH^{-} \\ksp = 4s^{3}$

The value finally becomes,

$ksp = 5.15 * x=10^{-5}$

<h3>What are the properties of Saturated solution ?</h3>

A saturated solution is one that contains all of the solute that can possibly dissolve in it. The most sodium chloride that may dissolve in $100 g$ of water at $20$ °C is $36.0 g$ . Past that point, adding more $NaCl$ won't cause it to dissolve because the solution is already saturated.

The solution reaches a point when it is saturated. This indicates that if you add more of the substance, it will stop dissolving and instead stay solid. The amount depends on how the solvent and solute interact molecularly.

To learn more about saturated solution, visit

brainly.com/question/9414660

#SPJ4

5 0

1 year ago

A 1.10 g sample contains only glucose and sucrose. When the sample is dissolved in water to a total solution volume of 25.0L, th

olga_2 [115]

Answer:

$\large \boxed{79 \, \%}$

Explanation:

I assume the volume is 2.50 L. A volume of 25.0 L gives an impossible answer.

We have two conditions:

(1) Mass of glucose + mass of sucrose = 1.10 g

(2) Osmotic pressure of glucose + osmotic pressure of sucrose = 3.78 atm

Let g = mass of glucose

and s = mass of sucrose. Then

g/180.16 = moles of glucose, and

s/342.30 = moles of sucrose. Also,

g/(180.16×2.50) = g/450.4 = molar concentration of glucose. and

s/(342.30×2.50) = s/855.8 = molar concentration of sucrose.

1. Set up the osmotic pressure condition

Π = cRT, so

$\begin{array}{rcl}\Pi_{\text{g}} +\Pi_{\text{s}}&=&\Pi_{\text{tot}}\\\dfrac{g}{450.4}\times8.314\times298 + \dfrac{s}{855.8}\times8.314\times298 & = & 3.78\\\\5.501g + 2.895s & = & 3.78\\\end{array}$

Now we can write the two simultaneous equations and solve for the masses.

2. Calculate the masses

$\begin{array}{lrcl}(1)& g + m & = & 1.10\\(2) &5.501g +2.895s & = & 3.78\\(3) & m & = &1.10 - g\\&5.501g + 2.895(1.10 - g) & = & 3.78\\&2.606g + 3.185 & = & 3.78\\ &2.606g & = & 0.595\\(4) & g & = & \mathbf{0.229}\\&0.229 + s & = & 1.10\\& s & = & \mathbf{0.871}\\\end{array}$

We have 0.229 g of glucose and 0.871 g of sucrose.

3. Calculate the mass percent of sucrose

$\text{Mass percent} = \dfrac{\text{Mass of component}}{\text{Total mass}} \times \, 100\%\\\\\text{Percent sucrose} = \dfrac{\text{0.871 g}}{\text{1.10 g}} \times \, 100\% = 79 \, \%\\\\\text{The mixture is $\large \boxed{\mathbf{79 \, \%}}$ sucrose}$

6 0

3 years ago