All categories

2 years ago

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]2 years ago

5 0

Answer: This is known as substitution cipher.

Explanation:

You might be interested in

41 to

tia_tia [17]

Answer:

d = 0.9 g/L

Explanation:

Given data:

Number of moles = 1 mol

Volume = 24.2 L

Temperature = 298 K

Pressure = 101.3 Kpa (101.3/101 = 1 atm)

Density of sample = ?

Solution:

PV = nRT (1)

n = number of moles

number of moles = mass/molar mass

n = m/M

Now we will put the n= m/M in equation 1.

PV = m/M RT (2)

d = m/v

PM = m/v RT ( by rearranging the equation 2)

PM = dRT

d = PM/RT

The molar mass of neon is = 20.1798 g/mol

d = 1 atm × 20.1798 g/mol / 0.0821 atm. L/mol.K × 273K

d = 20.1798 g/22.413 L

d = 0.9 g/L

4 0

2 years ago

What is wrong with this chemical equation - N2 + O2 → 2N20,

LUCKY_DIMON [66]

Answer and Explanation:

I think the thing that is wrong with this chemical equation is that there is 0 by the 2 instead of the letter O.

Other than that, everything would be balanced.

#teamtrees #PAW (Plant And Water)

5 0

2 years ago

Read 2 more answers

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

7 0

2 years ago

¿Que atomo gana el electron?

antoniya [11.8K]

Answer:

anion

Explanation:

3 0

2 years ago

A sample of iron has the dimensions of 2 cm x 3 cm x 2 cm. If the mass of this

77julia77 [94]

Answer: Volume iron = 2 x 3 x 2 => 12 cm³

Mass = 94 g

D = m / V

D = 94 / 12

D = 7.833 g/cm³

Explanation:

5 0

2 years ago