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Hoochie [10]
3 years ago
13

Is Pluto still a planet ? Explain your answer .

Chemistry
2 answers:
Ipatiy [6.2K]3 years ago
8 0

Hey!


According to scientists across the globe, Pluto is not a "planet", but a "dwarf planet." This also is determined when you were born and taught different ideals, cause everyone changed their sort of learning on planets overtime in our generation. Hope this helped!

iVinArrow [24]3 years ago
5 0

Hey,

I hope you are having a wonderful day

The Correct Answer is :

No Pluto Is not a Planet But It Is A Dwarf Planet

Explanation :

In 2006 The Definition Of the term "Planet" was changed Pluto was classified  to not be a planet but Instead A Dwarf Planet

 Hope This Helps :)

    - Sprinkles

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What type of molecule is the compound shown in the diagram? *diagram included*
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Monosaccharides are the simplest carbohydrates. Although glucose and fructose have the same molecular formula they have different structures. They cannot be further hydrolyzed to simple sugars. Disaccharides contains two monosaccharides. For example, lactose and sucrose. Polysaccharides on the other hand contains a large number of saccharides. An example is starch, glycogen and dextrans. Amino acids contains an amino acid, carboxyl group and an R-group. Whatever the diagram you have, you just look at the structures contained.
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PLEASE HELP CHEMISTRY QUESTION (picture attached)
Anastasy [175]

Answer:

It is called a salt bridge

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3 years ago
When a 10 mL graduated cylinder is filled to the 10 mL mark, the mass of the water was measured to be 10.085 g at 20ºC. If the d
Tanzania [10]
To calculate percent errorsubtract the accepted value from the experimental value.Take the absolute value of step 1.Divide that answer by the accepted value.Multiply that answer by 100
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5 0
3 years ago
Iron-59 is a beta emitter with a half-life of 44.5 days. If a sample initially contains 180 mg of iron-59, how much iron-59 is l
Jobisdone [24]

Answer:

2.79 mg iron-59 is left in the sample after 267 days.

Explanation:

Given that:

Half life = 44.5 days

t_{1/2}=\frac {ln\ 2}{k}

Where, k is rate constant

So,  

k=\frac {ln\ 2}{t_{1/2}}

k=\frac {ln\ 2}{44.5}\ days^{-1}

The rate constant, k = 0.0156 days⁻¹

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration  = 180 mg

Time = 267 days

So,  

[A_t]=180\ mg\times e^{-0.0156\times 267}

[A_t]=180\times e^{-0.0156\times 267}\ mg

[A_t]=2.79\ mg

<u>2.79 mg iron-59 is left in the sample after 267 days.</u>

5 0
3 years ago
Calculate the volume that a 0.323-mol sample of a gas will occupy at 265 K and a pressure of 0.900 atm. 4.36 L 4.36 L 4.63 L 4.6
ikadub [295]

The volume that a 0.323-mol sample of a gas will occupy at 265 K and a pressure of 0.900 atm is 7.81L.

<h3>HOW TO CALCULATE VOLUME?</h3>

The volume of a gas can be calculated using the following formula:

PV = nRT

Where;

  • P = pressure (atm)
  • V = volume (L)
  • R = gas law constant (0.0821 Latm/molK)
  • n = number of moles
  • T = temperature (K)

V = 0.323 × 0.0821 × 265 ÷ 0.900

V = 7.027 ÷ 0.9

V = 7.81L

Therefore, the volume that a 0.323-mol sample of a gas will occupy at 265 K and a pressure of 0.900 atm is 7.81L.

Learn more about volume at: brainly.com/question/1578538

4 0
2 years ago
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