Which of the following shows how rate depends on concentrations of reactants?

The process that makes atmospheric nitrogen available to plants by mutualistic and free-living bacteria is called

artcher [175]

Nitrogen fixation is the process that makes atmospheric nitrogen available to plants by mutualistic and free-living bacteria. The process is undertaken by the rhizobium bacteria that live in root roots of plants such as legumes. The mutualistic relationship is that the plant supplies the bacteria with a habitat in which to live, water, and nutrients, and the bacteria supply nitrogen for making plant proteins.

3 0

4 years ago

To convert from °F to °C: T(°C) = T(°F - 32) × 5/9

DaniilM [7]

Answer:

Your notation is a bit confusing, let me write it more clearly.

Explanation:

( Temperature in °F − 32) × 5/9 = Temperature in °C

4 0

3 years ago

URGENT!!!

miv72 [106K]

<h3>Answer;</h3>

= 13.629 g of ZnCl2

<h3><u>Explanation;</u></h3>

The equation for the reaction.

Zn(s)+2HCL(aq)=>ZnCl2(aq)+H2(g)

Number of moles of Zinc;

Moles = mass/RAM

= 6.5 g/65g/mol

= 0.1 moles

The mole ration of Zn : ZnCl2 is 1 : 1

Therefore, number of moles of ZnCl2 is 0.1 moles

Mass = moles × Molar mass

= 0.1 ×136.286 g/mol

<u> = 13.629 g</u>

7 0

3 years ago

Which combination of elements below will bond ionically and in the same ratio as lithium when it bonds to sulfur?

CaHeK987 [17]

<span>the answer is B) <span>sodium and oxygen </span></span>

3 0

3 years ago

Read 2 more answers

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