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Anastasy [175]
3 years ago
15

In an effort to try winning World War I, many countries broke previously created treaties and pacts that banned the use of chemi

cal-based weapons.
Based on the excerpt, how did chemical warfare advance during World War I?

a) Allied powers set bundles of wood and sulfur (S) on fire to smoke their enemies out of the trenches.

b) Production of ammonia for fertilizer was re purposed to cause enemies in the trenches to scatter because of the smell it produced.

c) Scientists were forced to stop their research on specific chemicals in order to fight in World War I.

d) Lethal gases, such as chlorine (Cl), were experimented with and used on enemy troops.

d?
Chemistry
1 answer:
nadya68 [22]3 years ago
3 0

D the belguim people atacked using it

You might be interested in
#8 please explain if you can<br> thanks
pav-90 [236]
The answer is D because the air is made of nitrogen and oxygen. The reaction is endothermic.
Hope this helps you! :) 

6 0
3 years ago
Keiko needs 100mL of a 5% acid solution for a science experiment. She has available a 1% solution and a 6% solution. How many mi
4vir4ik [10]

Answer:

Keiko should mix 20 mL 1% solution and 80 mL 6% solution for to make 100 mL 5% solution

Explanation:

There are 2 unknown values X= mL 6% solution and Y=1% solution. So, we need 2 equations:

1. Equation acid concentration. X mL 6% + Y mL 1% = 100 mL 5%

2. Equation solvent concentration X mL 94% + Y mL 99% = 100 mL 95%

When clearing X and Y :

(X mL 6%    +   Y mL 1%      =    100 mL 5%) (-15,7)

X mL 94%  +   Y mL 99%   =   100 mL 95%

_______________________________

 -                      Y  0.83       =   16.5

                        Y                =   19.9 mL 1% solution

Replace Y in anyone equation and X = 80 mL 6% solution

I hope to see been helpful

7 0
3 years ago
Read 2 more answers
Determine the root-mean square speed of CO2 molecules that have an average kinetic energy of 4.2 x10-21 J per molecule.
strojnjashka [21]
<span>Determine the root-mean-square sped of CO2 molecules that have an average Kinetic Energy of 4.21x10^-21 J per molecule. Write your answer to 3 sig figs.
</span><span>
E = 1/2 m v^2 

If you substitute into this formula, you will get out the root-mean-square speed. 

If energy is Joules, the mass should be in kg, and the speed will be in m/s. 

1 mol of CO2 is 44.0 g, or 4.40 x 10^1 g or 4.40 x 10^-2 kg. 

If you divide this by Avagadro's constant, you will get the average mass of a CO2 molecule. 

4.40 x 10^-2 kg / 6.02 x 10^23 = 7.31 x 10^-26 kg 

So, if E = 1/2 mv^2 

</span>v^2 = 2E/m = 2 (4.21x10^-21 J)/7.31 x 10^-26 kg = 115184.68 
Take the square root of that, and you get the answer 339 m/s.
8 0
3 years ago
A piece of unknown metal with mass 30 g is heated to 110.0 °C and dropped into 100.0 g of water at 20.0 °C. The final temperatur
Ymorist [56]

<u>Answer:</u> The specific heat of metal is 0.821 J/g°C

<u>Explanation:</u>

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]       ......(1)

where,

q = heat absorbed or released

m_1 = mass of metal = 30 g

m_2 = mass of water = 100 g

T_{final} = final temperature = 25°C

T_1 = initial temperature of metal = 110°C

T_2 = initial temperature of water = 20.0°C

c_1 = specific heat of metal = ?

c_2 = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

30\times c_1\times (25-110)=-[100\times 4.186\times (25-20)]

c_1=0.821J/g^oC

Hence, the specific heat of metal is 0.821 J/g°C

8 0
3 years ago
Explain why the electron configuration of 2-3-1 represents an atom in an excited state?
baherus [9]

Answer:

See explanation

Explanation:

If we look at the electron configuration closely, we will discover that the element must have had a ground state electron configuration of 2,4.

This is because, the innermost shell usually holds two electrons while the outer shells hold eight electrons each. The four electrons must be accommodated in the second shell in the ground state configuration of the compound.

However, when the atom is excited, one electron from this shell may move to the third shell to give the excited state configuration 2-3-1 as shown in the question.

6 0
3 years ago
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