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grandymaker [24]

3 years ago

9

PLZ PLZ PLZ HELPPPPPPPPPPPPP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

2 answers:

Vedmedyk [2.9K]3 years ago

8 0

Answer: c
explanation

Ganezh [65]3 years ago

4 0

Answer:

cold front

Explanation:

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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

Calculate the number of mol of solute in 8.00 × 102 mL of 0.250 M HCl.

Alenkasestr [34]

Moles = Volume*Molarity

= 0.008*0.250

= 0.002 mol

8 0

3 years ago

How much positive charge is in 1.4 kg of oxygen? The atomic weight (15.9994 g) of oxygen contains Avogadro’s number of atoms, wi

Cerrena [4.2K]

Answer:

$6.7511\times 10^7\ C$

Explanation:

The atomic weight of oxygen = 15.9994 g

This mass corresponds to 1 mole of the oxygen atoms.

Thus,

15.9994 g mass of oxygen contains $6.02\times 10^{23}$ atoms of oxygen.

1.4 kg = 1400 g ( 1 kg = 1000 g)

So,

1400 g mass of oxygen contains $\frac {6.02\times 10^{23}}{15.9994}\times 1400$ atoms of oxygen.

Number of atoms in 1400 g of oxygen = $526.769754\times 10^{23}$

Also, 1 atom of oxygen contains 8 protons

Charge of 1 proton = + $1.602\times 10^{-19}\ C$

So, Charge on 1 atom of oxygen = $8\times 1.602\times 10^{-19}\ C$

Thus,

Charge on $526.769754\times 10^{23}$ atoms of oxygen = $526.94476\times 10^{23}\times 8\times 1.602\times 10^{-19}\ C=6.7533\times 10^7\ C$

Thus, positive charge in 1.4 kg of oxygen = $6.7511\times 10^7\ C$

5 0

3 years ago

a paper clip has a length of 32.3mm. how many paper clips would you need to form a line of paper clips to cover a distance of 1.

nasty-shy [4]

Answer:

89883 paper clips

Explanation:

To solve this problem you need to use conversion factors to work with the same units.

First you should convert 32.3mm to miles:

$32.3mm*\frac{6.2*10^{-7}}{1000mm}=2*10^{-5}miles$

Then you should divide the total distance between the length of a paper clip in miles, so:

$\frac{1.8miles}{2*10^{-5}miles}=89883paperclips$

6 0

3 years ago

You do an experiment in which you need 0.25 mole of tyrosine (C,H11 NO3).

Marina CMI [18]

Answer:

B.90.5 g

Explanation:

not 100%sure hehe

6 0

3 years ago