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2 years ago

14

What could you do if you wanted to slow down the rate of a particular reaction?

1 answer:

hjlf2 years ago

5 0

Answer:

catalyst

Explanation:

this help toalter or speed up the rate of reaction could be the most to slow it

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How do crickets make sound from your their leg right?

Reptile [31]

The bottom of a cricket wing is covered with teeth-like ridges that make<span> it rough. The upper surface of the wing is like a scraper. When </span>crickets<span> rub the upper and lower parts of their wings together, they create a chirping</span>sound<span> called “stridulating."</span>

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3 years ago

Read 2 more answers

you are heating a solution and recording its temperature every 2 min during an experiment. which is your independent variable an

Zepler [3.9K]

An independent variable is the variable you are changing in order to measure the dependent variable, which is what you are measuring.

In this example, the

independent variable: chemicals in solution
dependent variable: temperature of solution

7 0

3 years ago

An atom has a diameter of 2.50 Å and the nucleus of that atom has a diameter of 9.00×10−5 Å . Determine the fraction of the volu

chubhunter [2.5K]

Answer:

The fraction of the volume of the atom that is taken up by the nucleus is $4.6656\times 10^{-14}$ .

The density of a proton is $6.278\times 10^{14} g/cm^3$ .

Explanation:

Diameter of the atom ,d = 2.50 Å

Radius of the atom ,r = 0.5 d=0.5 × 2.50 Å = 1.25Å

Volume of the sphere= $\frac{4}{3}\pi r^3$

Volume of atom = V

$V=\frac{4}{3}\pi r^3$ ..[1]

Diameter of the nucleus ,d' = $9.00\times 10^{-5}\AA$

Radius of the nucleus ,r' = 0.5 d'= $0.5\times 9.00\times 10^{-5}\AA=4.5\times 10^{-5}\AA$

Volume of nucleus = V'

$V=\frac{4}{3}\pi r'^3$ ..[2]

Dividing [2] by [1]

$\frac{V'}{V}=\frac{\frac{4}{3}\pi r'^3}{\frac{4}{3}\pi r^3}$

$=\frac{r'^3}{r^3}=\frac{(4.5\times 10^{-5}\AA)^3}{(1.25 \AA)^3}$

$\frac{V'}{V}=4.6656\times 10^{-14}$

The fraction of the volume of the atom that is taken up by the nucleus is $4.6656\times 10^{-14}$ .

Diameter of the proton ,d = $1.72\times 10^{-15} m = 1.72\times 10^{-13} cm$

1 m = 100 cm

Radius of the proton,r = 0.5 d= $0.5\times 1.72\times 10^{-13} cm=8.6\times 10^{-14} cm$

Volume of the sphere= $\frac{4}{3}\pi r^3$

Volume of atom = V

$V=\frac{4}{3}\times 3.14\times (8.6\times 10^{-14} cm)^3=2.664\times 10^{-39}cm^3$

Mass of proton, m = 1.0073 amu = $1.0073\times 1.66054\times 10^{-24} g$

$1 amu = 1.66054\times 10^{-24} g$

Density of the proton : d

$d=\frac{m}{V}=\frac{1.0073\times 1.66054\times 10^{-24} g}{2.664\times 10^{-39}cm^3}=6.278\times 10^{14} g/cm^3$

The density of a proton is $6.278\times 10^{14} g/cm^3$ .

5 0

3 years ago

40 POINTS COLLEGE CHEMISTRY

alexdok [17]

Answer:

C2= 0.16M

Explanation:

C1= 2M, V1= 20ml, C2= ?, V2= 250ml

Applying dilution formula

C1V1= C2V2

2×20 =C2×250

C2= 0.16M

5 0

3 years ago

Write the bond line formula for the compound (CH3)2CHCH2C(CH3)3.

maks197457 [2]

Formula is (CH3)2CHCH2C(CH3)3 and the name is 2,2,4-trimethylpentane

3 0

3 years ago