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horrorfan [7]
3 years ago
10

The simplest type of stream cipher, one in which one letter or character is exchanged for another, is known as what?

Chemistry
1 answer:
QveST [7]3 years ago
5 0

Answer: This is known as substitution cipher.

Explanation:

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Cho 6,72 gam Fe tan hết trong dung dịch H2SO4 đặc nóng dư thu được V lít khí SO2. Giá trị của V là
g100num [7]

4,032

BT electron

nSO2= 3nFe/

8 0
2 years ago
What kind of car do Cherry's and Marcia's boyfriends ride around in​
Bond [772]
I believe it was a blue mustang
6 0
3 years ago
Calculate the percent ionization of formic acid (hco2h) in a solution that is 0.311 m in formic acid and 0.189 m in sodium forma
ohaa [14]
<span>Answer: 0.094%


</span><span>Explanation:
</span>
<span></span><span /><span>
1) Equilibrium chemical equation:
</span><span />

<span>Only the ionization of the formic acid is the important part.
</span><span />

<span>HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq).
</span><span />

<span>2) Mass balance:
</span><span />

<span>                   HCOOH(aq)     HCOO⁻(aq)     H⁺(aq).

Start             0.311                 0.189

Reaction       - x                      +x                   +x

Final             0.311 - x          0.189 + x            x


3) Acid constant equation:
</span><span />

<span>Ka = [HCOO-] [N+] / [HCOOH] = (0.189 + x) x / (0.311 -x)
</span><span />

<span>= (0.189 + x )x / (0.311 - x) = 0.000177


4) Solve the equation:


You can solve it exactly (it will lead to a quadratic equation so you can use the quadratiic formula). I suggest to use the fact that x is much much smaller than 0.189 and 0.311.
</span><span />

<span>With that approximation the equation to solve becomes:


</span><span>0.1890x / 0.311 = 0.000177, which leads to:</span>
<span /><span>
x = 0.000177 x 0.311 / 0.189 = 2.91 x 10⁻⁴ M


5) With that number, the percent of ionization (alfa) is:
</span><span />

<span>percent of ionization = (moles ionized / initial moles) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (concentration of ions / initial concentration) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (0.000291 / 0.311)x 100 = 0.0936% = 0.094%
</span>
<span></span><span />
8 0
3 years ago
What is the concentration of magnesium bromide, in ppm, if 133.4 g MgBr2 dissolved in 1.84 L water. Then solve for the bromine c
Serhud [2]

Answer: 0.0725ppm

Explanation:

133.4g of MgBr2 dissolves in 1.84L of water.

Therefore Xg of MgBr2 will dissolve in 1L of water. i.e

Xg of MgBr2 = 133.4/1.84 = 72.5g

The concentration of MgBr2 is 72.5g/L = 0.0725mg/L

Recall,

1mg/L = 1ppm

Therefore, 0.0725mg/L = 0.0725ppm

7 0
3 years ago
Please help asap..........................
mrs_skeptik [129]

b or d

Explanation:

I think it's either b or d sorry if I'm wrong

7 0
3 years ago
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