All categories

1 year ago

Why is scientific notation used?

Chemistry

1 answer:

Setler [38]1 year ago

4 0

The correct option is (d) To express very large or very small number.

What is the Scientific Notation?

Scientific notation is a way to present numbers that are too large or too small to be easily written in decimal form. The three components of scientific notation are coefficient, base and exponent. The proper format to write a scientific notation is a x 10^b, where a is a number or decimal number and b is the power of 10 to make scientific notation equivalent to original number. When a number between 1 and 10 is multiplied by a power of 10 then the number is expressed in scientific notation. For example, 10000000 can be written as 10⁷, which is the scientific notation and the exponent is positive here. Similarly, for the negative exponent 0.000001 can be can be represented as 10-⁷. Hence, the scientific notation is used to express very large or very small numbers.

Learn more about Scientific Notation here: brainly.com/question/15361382

#SPJ1

You might be interested in

If we know this is a second-order reaction, what is the rate law?

Pavel [41]

Answer:

The rate is a mathematical relationship obtained by comparing reaction rate with reactant concentrations.

6 0

3 years ago

PLEASE HELP!!!

neonofarm [45]

Answer:

$Cl_2+2NaI\rightarrow I_2+2NaCl$

Explanation:

Hello there!

In this case, according to the described chemical reaction, Cl2 replaces iodine in NaI in order to produce I2 and NaCl:

$Cl_2+NaI\rightarrow I_2+NaCl$

It is possible to realize how chlorine replaces iodine in agreement with the single displacement reaction. Moreover, since chlorine and iodine atoms are not correctly balanced, we add a 2 in front of both NaI and NaCl in order to do so:

$Cl_2+2NaI\rightarrow I_2+2NaCl$

Best regards!

8 0

2 years ago

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Part (a) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

Part (b) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

5 0

3 years ago

The hydrogen-line emission spectrum includes a line at a wavelength of 434 nm. What is the energy of this radiation? (h= 6.626 x

Andrews [41]

Wavelength = 434nm = 434 x 10⁻⁹m
planck's constant = h= 6.626 x 10 ⁻³⁴ J
E =?
by using the formula;
E = hc /λ
value for c is 3 x 10⁸ m/s
E = (6.626 x 10 ⁻³⁴ J)(3 x 10⁸ m/s) / 434 x 10⁻⁹m
E = 1.9878 x 10⁻²⁵ / 434 x 10⁻⁹m
E = 4.58 x 10⁻¹⁹ joules

6 0

3 years ago

Scientists tracking the populations of certain freshwater fish have found an increase in the female to male ratio of several sma

vodomira [7]

I believe that the answer to the question provided above is that they formed two very different hypotheses since it is what they saw according to what they knew. Also hypothesis is just an intelligent guess.
Hope my answer would be a great help for you. If you have more questions feel free to ask here at Brainly.

7 0

3 years ago

Read 2 more answers