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ahrayia [7]
3 years ago
15

Why do the 8 phases of the moon occur

Chemistry
1 answer:
nadya68 [22]3 years ago
4 0
The phases of the moon occur because the moon orbits earth which causes the portion illuminated to change. The earth casts changing angles on the moon causing portions to be illuminated or dark.
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Draw the simplified curved arrow mechanism for the reaction of butan-2-one and CH3Li to give the major product.
Yuki888 [10]

Answer:

Follows are the solution to this question:

Explanation:

Please find the image file of the chemical reaction in the attachment:

In a water medium, the CH3- type CH 3Li is a heavy nucleophile that attacks the carbonyl carbon atom to form the alkoxide ion, which will then be protonated to form alcohol.

8 0
3 years ago
4) The initial rate of the reaction between substances P and Q was measured in a series of
ASHA 777 [7]

Answer:

The initial rate of the reaction between substances P and Q was measured in a series of

experiments and the following rate equation was deduced.

rate = k[P]^{2} [Q]

Complete the table of data below for the reaction between P and Q

Explanation:

Given rate of the reaction is:

rate= k[P]^{2} [Q]\\=>[Q]=\frac{rate}{k.[P]^{2} } \\and \\\\\\\ [P]=\sqrt{\frac{rate}{k.[Q]} }

Substitute the given values in this formulae to get the [P], [Q] and rate values.

From the first row,

the value of k can be calulated:

k=\frac{rate}{[P]^{2}[Q] } \\  =\frac{4.8*10^-3}{(0.2)^{2} 2. (0.30)} \\ =0.4

Second row:

2. Rate value:

rate =0.4* (0.10)^{2} * (0.10)\\\\        =4.0*10^-3mol.dm^-3.s^-1

3.Third row:

[Q]=\frac{rate}{k.[P]^{2} } \\     =9.6*10^-3 / (0.4 *(0.40)^{2} \\    =0.15mol.dm^{-3}

4. Fourth row:

[P]=\sqrt{\frac{rate}{k.[Q]} }\\=>[P]=\sqrt{\frac{19.2*10^-3}{0.60*0.4} } \\=>[P]=0.283mol.dm^{-3}

6 0
3 years ago
If m = 45g and V = 15ml, what is D in g/ml?​
ch4aika [34]

Answer:

3 g/mL

Explanation:

We know that the density of an object can be measured by dividing its mass (g) to its volume (mL).

Formula

D=m/v

Given data:

Mass= 45 g

Volume= 15 mL

Now we will put the values in formula:

D=45 g/ 15 mL= 3 g/mL

6 0
3 years ago
Suppose Gabor, a scuba diver, is at a depth of 15m15m. Assume that: The air pressure in his air tract is the same as the net wat
alexandr1967 [171]

Answer:

The ration of molar concentration is "2.5".

Explanation:

The given values are:

Average density of salt water,

= 1.03 \ g/cm^3

Net pressure,

= 2.00 \ atm

Increase in pressure,

= 1.00 \ atm

Now,

The under water pressure will be:

=  \frac{15 \ m}{10}\times 1 \ atm +1 \ atm

=  1.5\times 1+1

=  1.5+1

=  2.5 \ atm

hence,

The ratio will be:

=  \frac{(\frac{n}{V})_{15m} }{(\frac{n}{V})_{surface} }

or,

=  \frac{P}{P_s}

=  \frac{2.5}{1}

=  2.5

7 0
3 years ago
0.20dm
Alex_Xolod [135]

Answer:

I don't no sorry. ਕਿਉੁਂਕਿ ਸਤਿ ਸਤਿ ਸਤਿ ਸਤਿ ਸਤਿ ਸਤਿ ਸਤਿ ਸਤਿ ਸਤਿ ਸਤਿ ਸਤਿ ਸਤਿ ਸਤਿ ਸਤਿ ਸਤਿ ਸਤਿ 478figxigxurRyyv ਸ੍

5 0
3 years ago
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