All categories

3 years ago

Significant Figures: How many significant figures are in 865,010?

Chemistry

1 answer:

Harrizon [31]3 years ago

7 0

Answer: There are five significant figures in 865,010.

Explanation:

When a degree of accuracy is stated by each digit present in a mathematical figure then it is called a significant figure.

Rules for counting significant figures is as follows.

Any non-zero digits and zeros present between a non-zero figure are counted. For example, 3580009 has seven significant figures.

Trailing zeros are counted in a non-zero figure. For example, 0.00250 has three significant figures.

Leading zeros are not counted. For example, 0.0025 has two significant figures.

So, in the given figure 865010 has five significant figures and the trailing zero will not be counted.

Thus, we can conclude that there are five significant figures in 865,010.

You might be interested in

he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c

Leya [2.2K]

The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

<u>Answer:</u> The rate constant at 324°C is $61.29M^{-1}s^{-1}$

<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

$\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_{244^oC}$ = equilibrium constant at 244°C = $6.7M^{-1}s^{-1}$

$K_{324^oC}$ = equilibrium constant at 324°C = ?

$E_a$ = Activation energy = 71.0 kJ/mol = 71000 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = $244^oC=[273+244]K=517K$

$T_2$ = final temperature = $324^oC=[273+324]K=597K$

Putting values in above equation, we get:

$\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}$

Hence, the rate constant at 324°C is $61.29M^{-1}s^{-1}$

8 0

4 years ago

2.50 L of a gas at standard temperature and pressure is compressed to 575 mL. What is the new pressure of the gas

BartSMP [9]

Answer:

reek

Explanation:

5 0

3 years ago

What important material is absorbed by your digestive system besides water

Leto [7]

Answer:

The monosaccharides, amino acids, bile salts, vitamins, and other nutrients are absorbed by the cells of the intestinal lining

Explanation:

3 0

3 years ago

Read 2 more answers

Convert the following mole of ratio of elements to empirical formula 2.0 mol of Fe and 2.66 mol of O

Pavlova-9 [17]

Your answer = 3.2 .. thank you

3 0

3 years ago

For the following systems at equilibrium C: CaCO3(s) ⇌ CaO(s)+CO2(g) ΔH=+178 kJ/mol D: PCl3(g)+Cl2(g) ⇌ PCl5(g) ΔH=−88 kJ/mol cl

Rama09 [41]

Explanation:

C: $CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)$ ΔH=+178 kJ/mol

For an endothermic reaction, heat is getting absorbed during a chemical reaction and is written on the reactant side.

$A+\text{heat}\rightleftharpoons B$

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle. This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

Treat heat as a reactant and on increasing a reactant at equilibrium, shifts the reaction in the forward direction.

Increase temperature → increase in heat → forward direction

Decrease temperature → decease in heat → backward direction

System C - Increase temperature : Reaction will move forward

System C - Decrease temperature : Reaction will move backward

D: $PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$ ΔH=−88 kJ/mol

The total enthalpy of the reaction comes out to be negative .

The temperature of the surrounding will increase.

For an exothermic reaction, heat is released during a chemical reaction and is written on the product side.

$A\rightleftharpoons B+\text{ heat}$

Treat heat as a product and on increasing a product at equilibrium, shifts the reaction in the backward direction.

Increase temperature → increase in heat → backward direction

Decrease temperature → decease in heat → forward direction

System D - Increase temperature : Reaction will move backward

System D - Decrease temperature : Reaction will move forward

7 0

3 years ago