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SVETLANKA909090 [29]
4 years ago
9

Use the following graph of a car traveling on a straight northerly path to answer this question. At what time would the

Chemistry
1 answer:
BARSIC [14]4 years ago
6 0

Answer:

B 144.0 s is the best answer of this question

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When forming a contraction between "does" and "not," what's the correct way to write it? A. Don't B. Does't C. Doesn't D. Doesnt
KatRina [158]
C.Doesn't
do+not=Don't
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8 0
3 years ago
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Calculate the number of Cl atoms that are in 1.1 x 10^-21 moles of Cl atoms.
Margarita [4]

According to Avogadro's number there are 62.53 atoms of chlorine in 1.1x 10^-21 moles of Cl atoms.

<h3>What is Avogadro's number?</h3>

Avogadro's number is defined as a proportionality factor which relates number of constituent particles with the amount of substance which is present in the sample.

According to the definitions, Avogadro's number depend on determined value of mass of one atom of those elements.It bridges the gap between macroscopic and microscopic world by relating amount of substance with number of particles.

Number of atoms can be calculated using Avogadro's number as follows: mass/molar mass×Avogadro's number or number of moles×Avogadro's number.

ON substitution in above formula number of atoms=1.1×10²¹×6.023×10²³=62.53 atoms

Thus, there are 62.53 chlorine  atoms in 1.1x 10^-21 moles of Cl atoms.

Learn more about Avogadro's number,here:

brainly.com/question/11907018

#SPJ1

6 0
1 year ago
What is the correct formula (IUPAC) for sodium iodide
Shtirlitz [24]

Answer:

The correct formula is...

Nal

6 0
3 years ago
Explain why atoms move at different speeds depending on whether they are in liquid or solid
Diano4ka-milaya [45]
Because in difrent materials atoms are more compact or less compact.if they are less compact then it will be easear for them to move

8 0
4 years ago
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A solid piece of aluminum (51.0 g) was added to a solution of sodium hydroxide (84.1 g) in water, A balanced equation for this r
EastWind [94]
M_{Al}=26,98\frac{g}{mol}\\&#10;m=51g\\\\&#10;n=\frac{m}{M_{Al}}=\frac{51g}{26,97\frac{g}{mol}}\approx1,89mol\\\\\\&#10;M_{NaOH}=39,4\frac{g}{mol}\\&#10;m=84,1g\\\\&#10;n=\frac{m}{M_{NaOH}}=\frac{84,1g}{39,4\frac{g}{mol}}\approx2,14mol

<span>2 NaOH(aq)+ 2 Al(s)+ 2 H</span>₂<span>O → 2 NaAlO</span>₂<span>(aq)+ 3 H</span>₂<span>(g)
</span>  2mol      :     2mol           :                                  3mol
2,14mol   :     1,89mol      :                                  2,835mol
remains         completely consumed
2,14-1,89=0,25mol


A) Al

B)  

M_{NaOH}=39,4\frac{g}{mol}\\&#10;n=0,25mol \Rightarrow \ \ \ m=n*M_{NaOH}=0,25mol*39,4\frac{g}{mol}=9,85g

C)
n=2,835mol\\&#10;M_{H_{2}}=2,02\frac{g}{mol} \Rightarrow \ \ \ m=n*M_{H_{2}}=2,835mol*2,02\frac{g}{mol} \approx 5,73g

7 0
3 years ago
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