All categories

3 years ago

How many significant figures are in the measurement 811.40 grams?

Chemistry

2 answers:

Llana [10]3 years ago

8 0

Answer:

There is 5 significant figures in 811.40 grams

Explanation:

The significant figures are units that can be taking into account to write a figure.

Exists two general rules for it:

1. The numbers including zero (0) that are to the right of the figure are counted as a significant figure example: 345.60 has 5 significant figures

2. The zero (0) that are to the left of the figure are NOT counted as a significant figure example: 0.005 has 1 significant figure

igomit [66]3 years ago

3 0

There are 5 significant figures in this number because the zero counts and it comes after a number and on the right of the decimal point

You might be interested in

In solid NaCl, the equilibrium separation between neighboring Na+ and Cl- ions is 0.283 nm. Calculate the coulombic energy betwe

const2013 [10]

Explanation:

It is given that r = 0.283 nm. As 1 nm = $10^{-9} m$ .

Hence, 0.283 nm = $0.283 \times 10^{-9} m$

Formula for coulombic energy is as follows.

$U_{coulomb} = -1.748 \frac{e^{2}}{4 \pi \epsilon_{o} r}$

where, e = $1.6 \times 10^{-19}$ C

$\epsilon_{o}$ = $8.85 \times 10^{-12}$

$U_{coulomb} = -1.748 \frac{(1.6 \times 10^{-19}^{2}}{4 \times 3.14 \times 8.85 \times 10^{-12} \times 0.283 \times 10^{-9}}$

= $1.423 \times 10^{-18} J$

As 1 eV = $1.6 \times 10^{-19} J$

So, 1 J = $\frac{1 eV}{1.6 \times 10^{-19}}$

Hence, U = $\frac{1.423 \times 10^{-18} J}{1.6 \times 10^{-19} J}$

= 8.9 eV

Also, 1 J = $\frac{10^{-3} kJ}{6.022 \times 10^{23}mol}$

= $1.67 \times 10^{-27}$ kJ/mol

Therefore, U = $1.423 \times 10^{-18} J \times 1.67 \times 10^{-27}$ kJ/mol

= $2.37 \times 10^{-45} kJ/mol$

7 0

3 years ago

What is the volume of the sample at stp?

Yuki888 [10]

At stp the volume is 22.4 L .

hope this helps!

3 0

3 years ago

Definition: this is information from an experiment that can be assigned a number or a quantity and can often be used to construc

Contact [7]

The answer is “data”

8 0

3 years ago

Read 2 more answers

The atomic radius of a gold atom is 144×10^-12. The volume of a gold atom can be calculated using the volume of a sphere. The de

Alekssandra [29.7K]

Given:
Atomic radius of gold = 144 x 10⁻¹² m = 144 pm
Density of gold = 19.3 g/cm³ = 19.3 x 10⁻³ kg/cm³
Mass of gold sample = 1.40 g = 1.4 x 10⁻³ kg

Calculate the volume of an atom.
v = [(4π)/3]*(144 x 10⁻¹² m)³ = 1.2508 x 10⁻²⁹ m³

Calculate the mass of an atom.
$m=(1.2508\tmes10^{-29}\,m^{3})*(19.3\times 10^{-3}\, \frac{kg}{cm^{3}} )*(10^{2}\, \frac{cm}{m})^{3} \\ = 2.4142 \times 10^{-25} \, kg$

Calculate the number of atoms in a sample of mass 1.40 g.
$N = \frac{1.40 \time 10^{-3} kg}{2.4142 \times 10^{-25} \, \frac{kg}{atom} } =5.8 \times 10^{21} \, atoms$

Answer: 5.8 x 10²¹ atoms

3 0

3 years ago

What information do you get from the following

Gelneren [198K]

Explanation:

Kẽm tác dụng với axit clorua ( nhiệt độ ) thì sẽ tạo ra Kẽm II clorua và có khí Hidro bay lên

8 0

3 years ago