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Vinil7 [7]
3 years ago
14

An object is dropped from a height of 150m. What's the object's terminal velocity (the velocity with which the object hits the g

round)?
Chemistry
1 answer:
77julia77 [94]3 years ago
8 0
I used an online calculator and got 54.22 m/s. I hope that helps
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An unknown substance has been shown to have metallic bonds.
gladu [14]
This is the correct answer please mark me as brainliest

6 0
4 years ago
Read 2 more answers
A 2.5% (by mass) solution concentration signifies that there is of solute in every 100 g of solution. 2. therefore, when 2.5% is
densk [106]
Answer #1 is "there is 2.5 grams of solute in every 100 g of solution." 
We calculate for 2.5% by mass solution by dividing the mass of the solute by the mass of the solution and then multiply by 100.
Answer #2 is "that mass ratio would be 2.5/100 or 2.5 grams of solute/100 grams of solution." 
We weigh out 2.5 grams of solute and then add 97.5 grams of solvent to make a total of 100 gram solution, that is,
     mass of solute / mass of solution = 2.5g solute / (2.5g solute + 97.5g solvent)
                                                          = 2.5g solute / 100g solution
Answer#3 is "a solution mass of 1 kg is 10 times greater than 100 g, thus one kilogram (1 kg) of a 2.5% ki solution would contain 25 grams of ki."
We multiply 10 to each mass so that 100 grams becomes 1000grams since 1000 grams is equal to 1 kg:
     mass of solute / mass of solution = 2.5g*10/[(2.5g*10) + (97.5g*10)]
                                                          = 25g solute/(25g solute + 975g solvent)
                                                          = 25g solute/1000g solution
                                                          = 25g solute/1kg solution
5 0
3 years ago
Read 2 more answers
a compound has an empirical formula of CH2 what is the molecular formula if it's molar mass is 252.5 grams/mol
Goryan [66]

Empirical formula is the simplest ratio of components making up the compound. the molecular formula is the actual ratio of components making up the compound.

the empirical formula is CH₂. We can find the mass of CH₂ one empirical unit and have to then find the number of empirical units in the molecular formula.

Mass of one empirical unit - CH₂ - 12 g/mol x 1 + 1 g/mol x 2 = 12 = 14 g

Molar mass of the compound is - 252 .5 g/mol

number of empirical units = molar mass / mass of empirical unit

                                           = \frac{252.5 g/mol}{14 g}

                                           = 18 units

Therefore molecular formula is - 18 times the empirical formula

molecular formula  - CH₂ x 18 = C₁₈H₃₆                                            

molecular formula is C₁₈H₃₆  

8 0
3 years ago
The picture below is a great example of __________ biodiversity.
ss7ja [257]
Low because it is not diverse. It is just seed corn.
6 0
3 years ago
The reaction described by H2(g)+I2(g)⟶2HI(g) has an experimentally determined rate law of rate=k[H2][I2] Some proposed mechanism
MatroZZZ [7]

Answer:

Mechanism A and B are consistent with observed rate law

Mechanism A is consistent with the observation of J. H. Sullivan

Explanation:

In a mechanism of a reaction, the rate is determinated by the slow step of the mechanism.

In the proposed mechanisms:

Mechanism A

(1) H2(g)+I2(g)→2HI(g)(one-step reaction)

Mechanism B

(1) I2(g)⇄2I(g)(fast, equilibrium)

(2) H2(g)+2I(g)→2HI(g) (slow)

Mechanism C

(1) I2(g) ⇄ 2I(g)(fast, equilibrium)

(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)

(3) H(g)+I(g)→HI(g) (fast)

The rate laws are:

A: rate = k₁ [H2] [I2]

B: rate = k₂ [H2] [I]²

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]

<em>Where K' = K1 * K2</em>

C: rate = k₁ [H2] [I]

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]^1/2

Thus, just <em>mechanism A and B are consistent with observed rate law</em>

In the equilibrium of B, you can see the I-I bond is broken in a fast equilibrium (That means the rupture of the bond is not a determinating step in the reaction), but in mechanism A, the fast rupture of I-I bond could increase in a big way the rate of the reaction. Thus, just <em>mechanism A is consistent with the observation of J. H. Sullivan</em>

5 0
3 years ago
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