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timama [110]
2 years ago
12

You place a large pear on a scale and it shows a mass of 200 grams. What is the mass of that pear in centigrams? Remember: King

Henry Doesn’t Usually Drink Chocolate Milk! 2 centigrams 20,000 centigrams 2,000 centigrams 20 centigrams
Chemistry
1 answer:
statuscvo [17]2 years ago
7 0

Answer:

20,000 centigrams

Explanation:

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Calculate the number of moles in 25.0 g of each of the
mamaluj [8]

Answer:

A. 6.25moles

B. 0.78moles

C. 0.32moles

D. 0.15moles

E. 0.43moles

Explanation:

use

  • n = mass/molar mass

A. He

  • 25/4 = 6.25moles

B. O2

  • 25/32 = 0.78moles

C. Al(OH)3

  • 27+ 3(17) = 78
  • 25/78 = 0.32 moles

D. GaS3

  • 70+3(32) = 166
  • 25/166 = 0.15moles

E. C4H10

  • 4(12)+10(1) =58
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5 0
3 years ago
Need help plz asap .......​
olganol [36]

Answer:

Purple flowers

Explanation:

Usually the dominant allele is a capital letter.

From the question, purple flowers are dominant to white flowers while white are recessive.

PP = Purple

pp = White

4 0
3 years ago
Read 2 more answers
A particular first-order reaction has a rate constant of 1.35 × 102 s-1 at 25.0°c. what is the magnitude of k at 75.0°c if ea =
IgorC [24]
According to this formula:
K= A*(e^(-Ea/RT) when we have K =1.35X10^2 & T= 25+273= 298K &R=0.0821
Ea= 85.6 KJ/mol So by subsitution we can get A:
1.35x10^2 = A*(e^(-85.6/0.0821*298))
1.35x10^2 = A * 0.03
A= 4333
by substitution with the new value of T(75+273) = 348K & A to get the new K
∴K= 4333*(e^(-85.6/0.0821*348)
  = 2.16 x10^2
8 0
3 years ago
The term “ average atomic mass “ is a ______average so is calculator different Lee from a normal average
hoa [83]

Average atomic mass of an element is a sum of the product of the isotope mass and its relative abundance.

For example: Chlorine has 2 isotopes with the following abundances

Cl(35): Atomic mass = 34.9688 amu; Abundance = 75.78%

Cl(37): Atomic mass = 36.9659 amu; Abundance = 24.22 %

Average atomic mass of Cl = 34.9688(0.7578) + 36.9659(0.2422) =

                                             = 26.4993 + 8.9531 = 35.4524 amu

Thus, the term “ average atomic mass “ is a <u>weighted</u> average so it is calculated differently from a normal average

3 0
4 years ago
How many grams of O2 are required to produce 5 mol CO2
maxonik [38]
I did the math i got 220 grams
7 0
3 years ago
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