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padilas [110]
3 years ago
8

I need help with question 4 can y’all help me put them in the right spot Extra points !

Chemistry
1 answer:
mina [271]3 years ago
8 0
Tissue, cell, organ,organism, organ system
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A structure made up of two or more types of tissue is a
kakasveta [241]

Answer:

A structure made up of two or more types of tissue is an organ

Explanation:

Hope this helps

7 0
3 years ago
A salt forms in the reaction of barium with chlorine what is the most likely formula unit of this salt
Mashcka [7]
 Best Answer:<span>  </span><span>LiCl 

Cl and Li 

glucose 

BeF2 

last question answer 3 strong electrostatic requires a lot of energy
does this help though??</span>
3 0
3 years ago
A solution is prepared by dissolving 27.7 g of cacl2 in 375 g of water. the density of the resulting solution is 1.05 g/ml. the
bagirrra123 [75]
The answer is 6.88.
Solution:
We can calculate for the percent composition of CaCl2 by mass by dividing the mass of the CaCl2 solute by the mass of the solution and then multiply by 100. The total mass of the resulting solution is the sum of the mass of CaCl2 solute and the mass of water solvent. Therefore, the percent composition of CaCl2 by mass is 
     % by mass = (mass of the solute / mass of the solution)*100 
                        = mass of solute / (mass of the solute + mass of the solvent)*100
                        = (27.7 g CaCl2 / 27.7g + 375g) * 100 
                        = 6.88
5 0
3 years ago
Read 2 more answers
It takes 495.0 kJ of energy to remove 1 mole of electron from an atom on the surface of sodium metal. How much energy does it ta
Zigmanuir [339]

Answer:

\lambda=241.9\ nm

Explanation:

The work function of the sodium= 495.0 kJ/mol

It means that  

1 mole of electrons can be removed by applying of 495.0 kJ of energy.

Also,  

1 mole = 6.023\times 10^{23}\ electrons

So,  

6.023\times 10^{23} electrons can be removed by applying of 495.0 kJ of energy.

1 electron can be removed by applying of \frac {495.0}{6.023\times 10^{23}}\ kJ of energy.

Energy required = 82.18\times 10^{-23}\ kJ

Also,  

1 kJ = 1000 J

So,  

Energy required = 82.18\times 10^{-20}\ J

Also, E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

So,  

79.78\times 10^{-20}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}

\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{82.18\times 10^{-20}}

\lambda=\frac{10^{-26}\times \:19.878}{10^{-20}\times \:82.18}

\lambda=\frac{19.878}{10^6\times \:82.18}

\lambda=2.4188\times 10^{-7}\ m

Also,  

1 m = 10⁻⁹ nm

So,  

\lambda=241.9\ nm

6 0
3 years ago
ALGEBRA 1 SEM 1
olganol [36]

Answer:

This is the best answer your mom

Explanation:

6 0
2 years ago
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