All categories

2 years ago

How can we predict how the moon will change appearance from day to day?

Chemistry

2 answers:

kvv77 [185]2 years ago

7 0

Answer:

1. When you see the moon, think of the whereabouts of the sun

2. The moon rises in the east and sets in the west, each and every day

3. The moon takes about a month (one month) to orbit the Earth

4. The moon’s orbital motion is toward the east

Explanation:

OlgaM077 [116]2 years ago

3 0

Answer:

Explanation:

The slice of sunlight continues to decrease until the moon is a waning crescent and then a new Moon. The whole cycle (from new Moon to new Moon) takes about 29.5 days. If you have a hard time remembering which way the moon phases go, just think: “white on right, getting bright!”

You might be interested in

How many sodium ions are in the initial 50.00-mL solution of Na2CO3

tresset_1 [31]

From other sources, the given mass of the solute that is being dissolved here is 7.15 g Na2CO3 - 10H2O. We use this amount to convert it to moles of Na2CO3 by converting it to moles using the molar mass then relating the ratio of the unhydrated salt with the number of water molecules. And by the dissociation of the unhydrated salt in the solution, we can calculate the moles of Na+ ions that are present in the solution.

Na2CO3 = 2Na+ + CO3^2-

7.15 g Na2CO3 - 10H2O (1 mol / 402.9319 g) (1 mol Na2CO3 / 1 mol Na2CO3 - 10H2O) ( 1 mol Na2CO3 / 1 mol Na2CO3-10H2O ) ( 2 mol Na+ / 1 mol Na2CO3) = 0.04 mol Na+ ions present

8 0

2 years ago

Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by No

liraira [26]

Answer:

Molality = 0.066 m

Molar mass = 608.36 g/mol

Explanation:

It seems the question is incomplete. However a web search us shows this data:

" Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in 25.0 g camphor, the freezing-point depression is 2.63 °C (Kf for camphor is 40 °C·kg/mol). Calculate the molality of the solution and the molar mass of reserpine. "

The freezing-point depression is expressed by:

ΔT=Kf * m

We put the data given by the problem and solve for m:

2.63 °C = 40°C·kg/mol * m

m = 0.06575 m

For the calculation of the molar mass: Molality is defined as moles of solute per kilogram of solvent:

0.06575 m = Moles reserpine / kg camphor

25.0 g camphor ⇒ 25.0/1000 = 0.025 kg camphor

We calculate moles of reserpine:

0.06575 m = Moles reserpine / 0.025 kg camphor

Moles reserpine = 1.64x10⁻³ mol

Finally we use the mass of reserpine and the moles to calculate the molar mass:

1.00 g reserpine / 1.64x10⁻³ mol = 608.36 g/mol

Keep in mind that if the data in your problem is different, the results will be different. But the solving method remains the same.

8 0

2 years ago

6.74 g of the monoprotic acid KHP (MW = 204.2 g/mol) is dissolved into water. The sample is titrated with a 0.703 M solution of

Dennis_Churaev [7]

Answer:

Volume of the calcium hydroxide solution used is 0.0235 mL.

Explanation:

$2KHP+Ca(OH)_2\rightarrow 2H_2O+Ca(KP)_2$

Moles of KHP = $\frac{6.74 g}{204.2 g/mol}=0.0330 mol$

According to reaction, 2 moles of KHP with 1 mole of calcium hydroxide , then 0.0330 moles of KHP will recat with ;

$\frac{1}{2}\times 0.0330 mol=0.01650 mol$ of calcium hydroxide

Molarity of the calcium hydroxide solution = 0.703 M

Volume of calcium hydroxide solution = V

$Molarity=\frac{Moles}{Volume(L)}$

$0.703 M=\frac{0.01650 M}{V}$

$V=\frac{0.01650 M}{0.703 M}=0.0235 mL$

Volume of the calcium hydroxide solution used is 0.0235 mL.

4 0

2 years ago

Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit

Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

$0.0811 M=\frac{n}{0.0017 L}$

n = 0.0001378 mol

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

$\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol$ of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

$x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L$

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0

2 years ago

NO2 can react with the NO in smog, forming a bond between the N atoms. Draw the structure of the resulting compound, including f

ELEN [110]

First, let's write down the balanced chemical reaction between the given reactants:

NO₂ + NO → N₂O + O₂

The Lewis structure of the main product is shown in the attached picture. To determine the formal charge of each element, the formula is as follows:

Formal Charge = Valence electrons - Non-bonding valence electrons - (Bonding electrons/2)

For the leftmost N:
Formal charge = 5 - 2 - 6/2 = 0
For the middle N:
Formal charge = 5 - 0 - 8/2 = 1
For O:
Formal charge = 6 - 6 - 2/2 = -1

6 0

1 year ago