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3 years ago

Why does the moon appear to be so bright in the sky

Chemistry

2 answers:

Nana76 [90]3 years ago

6 0

Because of the sun reflects onto the moon and makes it brighter.

maria [59]3 years ago

5 0

Answer:

The shine of the Sun on the lunar surface brightens it, while also giving the moon "dark sides". Its position in Earth orbit allows it to have certain phases because of its orbital position relative to the sun. You're welcome.

Explanation:

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Analysis of skunk spray yields a molecule with 44.77% c, 7.46% h and 47.76% s by mass. what is the empirical formula for this mo

stiv31 [10]

Suppose we have 100 gr of the substance. Then by weight, it would contain 44.77 gr of C, 7.46 gr of H and 47.76 gr of S. We need to look up the atomic weights of these atoms; M_H=1, M_C=12, M_S=32. The following formula holds (where n are the moles of the substance, M its molecular mass and m its mass): n=m/M. Substituting the known quantities for each element, we get that the substance has 3.73 moles of C, 7.46 moles of H and 1.49 moles of S. In the empirical formula for the molecule, all atoms appear an integer amout of times. Hence, for every mole of Sulfur, we have 2.5 moles of C and 5 moles of H (by taking the moles ratios). Thus, for every 2 moles of sulfur, we have 5 moles of C and 10 moles of H. Now that all the coefficients are integer, we have arrived at an empirical formula for the skunk spray agent: $C_5H_{10}S_2$

4 0

3 years ago

Calculate the ph of a solution containing 0.0451 m potassium hydrogen tartrate and 0.028 m dipotassium tartrate. The ka values f

Marina86 [1]

Given buffer:

potassium hydrogen tartrate/dipotassium tartrate (KHC4H4O6/K2C4H4O6 )

[KHC4H4O6] = 0.0451 M

[K2C4H4O6] = 0.028 M

Ka1 = 9.2 *10^-4

Ka2 = 4.31*10^-5

Based on Henderson-Hasselbalch equation;

pH = pKa + log [conjugate base]/[acid]

where pka = -logKa

In this case we will use the ka corresponding to the deprotonation of the second proton i.e. ka2

pH = -log Ka2 + log [K2C4H4O6]/[KHC4H4O6]

= -log (4.31*10^-5) + log [0.0451]/[0.028]

pH = 4.15

4 0

3 years ago

Particulate Electrical Forces Unit Test for Connexus, Chemistry A

pickupchik [31]

I’m sorry no one helped you with those questions. But could you please be a doll and give me the answer for the multiple-choice questions 1-15. Pleaseee I’m so desperate.?

3 0

2 years ago

If exactly 59.6 g of nitrogen gas is needed to inflate your air bag to the

Inga [223]

Answer:

We need 92.3 grams of sodium azide

Explanation:

Step 1: Data given

Mass of nitrogen gas = 59.6 grams

Molar mass of nitrogen gas = 28.0 g/mol

Molar mass of sodium azide = 65.0 g/mol

Step 2: The balanced equation

2NaN3 → 2Na + 3N2

Step 3: Calculate moles nitrogen gas

Moles N2 = mass N2 / molar mass N2

Moles N2 = 59.6 grams/ 28.0 g/mol

Moles N2 = 2.13 moles

Step 4: Calculate moles NaN3

for 2 moles NaN3 we'll have 2 moles Na and 3 moles N2

For 2.13 moles N2 we need 2/3* 2.13 = 1.42 moles NaN3

Step 5: Calculate mass NaN3

Mass NaN3 = Moles NaN3 * molar mass NaN3

Mass NaN3 = 1.42 moles * 65.0 g/mol

Mass NaN3 = 92.3 grams

We need 92.3 grams of sodium azide

7 0

3 years ago

Calculate the pH of: (a) 0.1M HCl; (b) 0.1M NaOH; (c) 3 X 10% M HNO3; (d) 5 X 10-10 M HCIO.; and (e) 2 x 10-8 M KOH.

Shtirlitz [24]

Answer:

(a) $pH = -Log (0.1M) = 1$

(b) $pH = -Log (10^{-13}M) = 13$

(d) $pH = -Log (4.93x10^{-10}M) = 9.3$

(e) $pH = -Log (5^{-7}M) = 6.3$

Explanation:

To calculate de pH of an acid solution the formula is:

$pH = -Log ([H^{+}]) = 1$

were [H^{+}] is the concentration of protons of the solution. Therefore it is necessary to know the concentration of the protons for every solution in order to solve the problem.

(a) and (c) are strong acids so they dissociate completely in aqueous solution. Thus, the concentration of the acid is the same as the protons.

(b) and (e) are strong bases so they dissociate completely in aqueous solution too. Thus, the concentration of the base is the same as the oxydriles. But in this case it is necessary to consider the water autoionization to calculate the protons concentration:

$K_{w} =[H^{+} ][OH^{-}]=10^{-14}$