CARDS

Muiltplying 2.5 x 10^10 by 3.5 x 10^-7

andrey2020 [161]

Answer:

The value of $\left(2.5 \times 10^{10}\right) \times\left(3.5 \times 10^{-7}\right)$ is 8750

Solution:

$\left(2.5 \times 10^{10}\right) \times\left(3.5 \times 10^{-7}\right)$

$\Rightarrow\left(2.5 \times 10^{10} \times 3.5 \times 10^{-7}\right)$

$\Rightarrow\left(2.5 \times 3.5 \times\left(10^{10} \times 10^{-7}\right)\right)$

$\Rightarrow(2.5 \times 3.5) \times\left(10^{10+(-7)}\right)$

$\Rightarrow(2.5 \times 3.5) \times 10^{3}$

$\Rightarrow 8.75 \times 10^{3}$

$\Rightarrow 8.75 \times 1000=8750$

6 0

3 years ago

A) A metal ruler has a mass of 42g and a volume of 10cm". What is the density?

nataly862011 [7]

Answer:

4.2g/cm^3

Explanation:

Density is a measurement of the amount of mass of a substance can fit into a space. In order to find density you can always use the formula mass/volume (often abbreviated as m/v). In this case that would be 42/10 g/cm^3 or 4.2g/cm^3

3 0

4 years ago

PLS HELPPP How many moles of oxygen atoms (not molecules) are present in 6.41×1025 molecules of dinitrogen pentoxide (N2O5)? 1.

olga55 [171]

Answer:

(4) 266 moles

Explanation:

We have Dinitrogen Pentoxide N2O5

6.41*10^25 molecules are given

No of Moles of N2O5 = $\frac{No of Molecules }{Avogadro's Number}$

= $\frac{6.41 * 10^25}{6.022*10^23}$

= 106.5 mol

Now using Unitary Method

2 Mole of Nitrogen pentoxide require 5 mole of Oxygen to form N2O5

1 mole of N = $\frac{5}{2}*O$

In 106.5 mole of N = $\frac{5}{2}*106.5$ = 266.25 mole

So, 6.41*10^25 molecules of N2O5 will require 266.25 mole of Oxygen atoms.

5 0

3 years ago

What are the elements and number of atoms in the compound c6h12o6?

VARVARA [1.3K]

Carbon-6, Hydrogen-12, Oxygen-6
Number of atoms altogether: 24

3 0

3 years ago

Which activity most likely involves the direct use of saturated hydrocarbons?

Ugo [173]

I dont know but the options are:
burning natural gasdissolving salt in watercarving stone with hammer and chiselbending steel by hand

8 0

4 years ago

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