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Mumz [18]
3 years ago
15

Convert 239 km/hr into m/min

Chemistry
2 answers:
eduard3 years ago
6 0

Answer:

Explanation:

we know that 1 km =1000 m and 1 hour =60 min

therefore

239 km/hr =239*1000m/60 min=3983.3 m/min

ValentinkaMS [17]3 years ago
5 0

Answer:

3983.3333334 m/min

Explanation:

1 km/h is equal to 16.6667 m/min so 239 km/h is equal to 3983 .3334 m/min.

239(16.6667)= 3983.3334

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How would you male 2.0L of O.100M sulfurice acid from a 12.0M stock bottle of sulfurice acid Calc, sketch, and write a summary s
ozzi

<u>Answer:</u> The volume of stock solution needed is 0.016 L

<u>Explanation:</u>

To calculate the volume of stock solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the stock solution

M_2\text{ and }V_2 are the molarity and volume of diluted solution

We are given:

M_1=12M\\V_1=?L\\M_2=0.10M\\V_2=2L

Putting values in above equation, we get:

12\times V_1=0.1\times 2\\\\V_1=0.016L

Hence, the volume of stock solution needed is 0.016 L

7 0
3 years ago
How many moles of koh are required to prepare 0.80 l of 6.4 m koh? 0.13 mol koh 5.1 mol koh 7.2 mol koh 8.0 mol koh?
wel
Answer is: 5,1 mol KOH.
V (KOH) = 0,8l = 0,8 dm³. 
c (KOH) = 6,4 mol/dm³.
V - volume of solution.
c - concentration of solution.
c = n÷V
n - amount of substance.
n (KOH) = c·V = 6,4 mol/dm³ · 0,8 dm³ = 5,12 mol.
Potassium hydroxide (KOH) can<span> </span>dissolve<span> in water at room temperature.</span>


8 0
4 years ago
DNA can be replicated through an in vivo mechanism or a polymerase chain reaction (PCR) mechanism. Match each description to the
umka2103 [35]

Complete question from other source attached

Answer:

Explanation:

Catalyzed by DNA polymerase - both. DNA polymerase catalyzes DNA replication in the cell. However, purified versions of the enzyme are also used to synthesise DNA as part of PCR reactions

involves leading strand synthesis only - PCR. In PCR, lagging strand synthesis is not carried out because the DNA is denatured (rendered single stranded). Therefore, each strand is replicated independently by leading strand synthesis

duplicates a small fragment of the genome - PCR. Usually, to carry out PCR, small sequences called primers are used that specify the region of DNA to be replicated

duplicates the entire genome - in vivo replication - when the DNA is replicated in vivo, the entire genome is replicated. This is carried out prior to cell division so that two daughter cells can each inherit a copy of the entire genome

7 0
4 years ago
What is the rate law for the reaction 2A + 2B + 2C --&gt; products
-Dominant- [34]

Answer:

R = 47.19 [A]*([B]^2)*[C]

Explanation:

The rate law for the reaction 2A + 2B + 2C --> products

Is being sought.

The reaction rate R could be expressed as  

R = k ([A]^m)*([B]^n)*([C]^p)                      (1)

where m, n, and p are the reaction orders with respect to (w.r.t.) components A, B and C respectively. This could be reduced to

R = ka ([A]^m)                   (2)

Where ka=(k[B]^n)*([C]^p);    

R = kb ([B]^n)                    (3)

Where kb=(k[A]^m)*([C]^p); and  

R = kc ([C]^p)                     (4)

Where kc=(k[A]^m)*([B]^n).

Equations (2), (3) and (4) are obtained for cases when the concentrations of two components are kept constant, while only one component’s concentration is varied. We can determine the reaction wrt each component by employing these equations.  

The readability is very much enhanced when the given data is presented in the following manner:

Initial [A]  0.273   0.819   0.273   0.273

Initial [B]  0.763   0.763   1.526   0.763

Initial [C]  0.400   0.400    0.400   0.800

Rate           3.0       9.0       12.0       6.0

Run#  1  2 3  4

Additional row is added to indicate the run # for each experiment for easy reference.

First, we use the initial rate method to evaluate the reaction order w.r.t. each component [A], [B] and [C] based on the equations (2), (3) and (4) above.

Let us start with the order wrt [A]. From the given data, for experimental runs 1 and 2, the concentrations of reactants B and C were kept constant.

Increasing [A] from 0.273 to 0.819 lead to the change of R from 3.0 to 9.0, hence we can apply the relation based on equation (2) between the final rate R2, the initial rate R1 and the final concentration [A2] and the initial concentration [A1] as follows:

R2/R1=ka[A2]^m/ka[A1]^m=([A2]/[A1])^m

9.0/3.0 = (0.819/0.273)^m

3 = (3)^m = 3^1  -> m = 1

Similarly, applying experimental runs 1 and 3 could be applied for the determination of n, by employing equation (3):  

R3/R1=kb[B3]^n/kb[B1]^n=([B3]/[B1])^n

12/3= (1.526/0.763)^n

4= 2^n, -> n = 2

And finally for the determination of p we have using runs 4 and 1:

R4/R1=kc[C4]^p/kc[C1]^p=([C4]/[C1])^p

6/3= (0.8/0.4)^p

2= 2^p , -> p = 1

Therefore, plugging in the values of m, n and p into equation (1), the rate law for the reaction will be:

R = k [A]*([B]^2)*[C]

The value of the rate constant k could be estimated by making it the subject of the formula, and inserting the given values, say in run 1:

k = R /( [A]*([B]^2)*[C]) = 3/0.273*(0.763^2)*0.4 =

47.19

Finally, the rate law is

R = 47.19 [A]*([B]^2)*[C]

7 0
4 years ago
Compare and contrast Lamarck's and Darwin's theories of change over time.36a. What are the similarities? What are the difference
frosja888 [35]

Answer:

See explanation

Explanation:

Lanark proposed a theory of evolution hinged on use and disuse of body parts. This is a theory that upholds the transfer of acquired traits to offsprings. Darwin's theory of evolution was rather based on natural selection or 'survival of the fittest'. Darwin's theory has more evidence backing it compared to Lamarck's theory. The field of genetics which appeared long after Darwin's death confirmed that inherited traits are indeed passed from parents to offsprings. These inherited traits are usually those that give organisms an edge over others in their environment.

Earthworms are brown and not green or beige because, the brown earthworms have the same colour as the soil and are better enabled to escape from predators. From Darwin's theory, the brown earthworms are the 'fittest' hence they survive in their environment. Lamarck's theory may approach the matter from the point that earthworms move with their bodies on the soil more frequently, hence they 'acquired' the brown colour and passed it on to their offspring.

From the activity, "Hiding in the Background", individual worms spent their entire life cycle in either green or beige colour. Their colour didn't change as a result of the environment where they lived, rather the green earthworms were more in number, not particularly because of any specific changes the earthworms themselves had undergone, but because most of their predators concentrated on eating the beige worms. As a result of this, there were many green worms when the next generation began compared to the number of beige worms. Hence, there were more green earthworms.

8 0
4 years ago
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