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Marizza181 [45]
3 years ago
8

(WILL GIVE BRAINLESS)Throwing a paper airplane through the air is an example of A. fluid friction B. sliding friction C. rolling

friction D. solid friction ​
Chemistry
2 answers:
Scilla [17]3 years ago
8 0

Answer:

sliding friction

Explanation:

Ymorist [56]3 years ago
7 0

Answer:

D. solid friction

Explanation:

Its not a fluid, Its not sliding, and its not rolling, so therefor is solid friction

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2) Is an atom with 26 protons and 30 neutrons an element or an isotope?*<br> Element<br> O Isotope
NikAS [45]
Does the element have a different amount of electrons than protons?
8 0
2 years ago
Calculate the number of moles of ethanol acid in 30 cm³of 1 mol dm-³ solution ​
muminat

Answer:

May I assume "ethanol acid is just ethanol (it has one slightly acidic H atom).  If so, the molar mass is 46.02 g/mole.

Explanation:

We have 30 cm^3 [30 ml] of 1.0 M (1 mole/liter)  [1 dm³ = 1 liter].

That is 1 mole/liter.  30 ml would contain (0.030 liter)*(1 mole/1 liter) = 0.03 moles.

7 0
3 years ago
Help me please!! help would be appreciated
valina [46]
I don’t know if this is right but I thing It’s B
6 0
3 years ago
Read 2 more answers
An unknown aqueous metal analysis yielded a detector response of 0.255. When 1.00 mL of a solution containing 100.0 ppm of the m
AVprozaik [17]

Answer:

1.022ppm is the unknown concentration of the metal

Explanation:

Based on Lambert-Beer law, the increasing in signal of a detector is directly proportional to its concentration.

The unknown concentration (X) produces a signal of 0.255

99mL * X + 1mL * 100ppm / 100mL produces a signal of 0.502

0.99X + 1ppm produce 0.502, thus, X is:

0.255 * (0.99X + 1 / 0.502) =

X = 0.503X + 0.508

0.497X = 0.508

X =

1.022ppm is the unknown concentration of the metal

3 0
3 years ago
34g aluminum are combined with 39g chlorine gas which is the limiting reactant
vampirchik [111]

The limiting reactant is chlorine (Cl2).

<u>Explanation</u>:

Limiting reactant is the amount of product formed which gets limited by the reagent without continuing it.

        2 Al + 3 Cl2 ==> 2 AlCl3  represents the balanced equation.

Number of moles Al present = 34 g Al x 1 mole Al / 26.98 g

                                                = 1.260 g moles of Al

Number of moles Cl2 present = 39 g Cl2 x 1 mole Cl2 / 35.45 g

                                                  = 1.10 g moles of Cl2

Dividing each reactant by it's coefficient in the balanced equation obtains:

1.260 moles Al / 2 = 0.63 g moles of Al

1.11 moles Cl2 / 3 = 0.36 g moles of Cl2

The reactant which produces a lesser amount of product is called as limiting reactant.

Here the Limiting reactant is Cl2.

6 0
3 years ago
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