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zvonat [6]
3 years ago
12

Lori is saving money to buy a new bike. She needs $120 but has only saved 60%.

Chemistry
1 answer:
I am Lyosha [343]3 years ago
3 0
10% of 120 is 12
12 x 6 is 72
The answer is 72
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Compare the root mean square speeds of O2 and UF6 at 65 degree Celsius
gayaneshka [121]
<h3>Answer:</h3>

The root mean square speeds of O₂ and UF₆ is 513m/s and 155 m/s respectively.

<h3>Solution and Explanation:</h3>
  • To find how fast molecules or particles of gases move at a particular temperature, the root  mean square speed is calculated.
  • Root mean square speed of a gas is calculated by using the formula;

       Root mean square=\sqrt{ \frac{3RT}{M}

       Where R is the molar gas constant, T is the temperature and M is the molar mass of gas in Kg.

<h3>Step 1: Root mean square speed from O₂</h3>

Molar mass of Oxygen is 32.0 g/mol or 0.032 kg/mol

Temperature = 65 degrees Celsius or 338 K

Molar gas constant = 8.3145 J/k.mol

Root mean square speed = \sqrt\frac{(3)(8.3145)(338K)}{0.032}

                                        = 513.289 m/s

<h3>Step 2: Root mean square speed of UF₆   </h3>

The molar mass of UF₆ is 352 g/mol or 0.352 kg/mol        

Root mean square speed = \sqrt\frac{(3)(8.3145)(338K)}{0.352}

                                                 = 154.762m/s

Therefore; the root mean square speeds of O₂ and UF₆ is 513m/s and 155 m/s respectively.

7 0
3 years ago
Calculate the amount in grams of Na2CO3 needed in a reaction with HCl to produce 120g NaCl?
klemol [59]
The balanced chemical reaction is written as :

Na2CO3<span> + 2HCl === 2NaCl + H2O + CO2
</span>
We are given the amount of NaCl to be produced from the reaction. This will be the starting point for the calculations. We do as follows:

120 g NaCl ( 1 mol / 58.44 g) ( 1 mol Na2CO3 / 2 mol NaCl)( 105.99 g / 1 mol ) = 1108.82 g Na2CO3 needed
8 0
3 years ago
A certain radioactive nuclide has a half life of 1.00 hour(s). Calculate the rate constant for this nuclide. s-1 Calculate the d
Karo-lina-s [1.5K]

Answer:

k= 1.925×10^-4 s^-1

1.2 ×10^20 atoms/s

Explanation:

From the information provided;

t1/2=Half life= 1.00 hour or 3600 seconds

Then;

t1/2= 0.693/k

Where k= rate constant

k= 0.693/t1/2 = 0.693/3600

k= 1.925×10^-4 s^-1

Since 1 mole of the nuclide contains 6.02×10^23 atoms

Rate of decay= rate constant × number of atoms

Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms

Rate of decay= 1.2 ×10^20 atoms/s

8 0
3 years ago
A car with a mass of 1,500 kg is traveling at a speed of 30 m/s. What force must be applied to stop the car in 3 seconds ?
saw5 [17]

Yo sup??

we can solve this problem by applying Newton's 2nd law

F*t=Δp

p=momentum

pi=mu=1500*30

pf=mv=m*0=0

Therefore

F*3=1500*30

F=15000 N

Hope this helps.

5 0
3 years ago
KCI is a molecule. True or False
Monica [59]

Answer: duh

Explanation:

4 0
3 years ago
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