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Nesterboy [21]
3 years ago
13

The PC modem was invented in what year?​

Chemistry
1 answer:
Andrew [12]3 years ago
5 0

Answer:

1977

Explanation:

Getting connected: a history of modems. Then, in 1977, Dale Heatherington and Dennis Hayes created the world's first PC modem, the 80-103A.

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4. Which is produced when a base reacts with water?
Dennis_Churaev [7]

Answer:

hydroxide ion

Explanation:

I took the quiz

5 0
3 years ago
Read 2 more answers
How do we balance Zn + HNO3 Zn(NO3)2 + NO + H20​
USPshnik [31]

3Zn + 8HNO₃⇒ 3Zn(NO₃)₂ + 2NO + 4H₂O

<h3>Further explanation </h3>

Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:  

  • 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.  
  • 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product  
  • 3. Select the coefficient of the substance with the most complex chemical formula equal to 1  

For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:  

Balancing C atoms, H and the last O atoms

Reaction

Zn + HNO₃⇒ Zn(NO₃)₂ + NO + H₂O

  • 1. gives a coefficient

aZn + bHNO₃⇒ Zn(NO₃)₂ + cNO + dH₂O

  • 2. make an equation

Zn : left = a, right =1 ⇒a=1

H : left = b, right = 2d⇒ b=2d (eq 1)

N : left = b, right = 2+c⇒b=2+c (eq 2)

O : left = 3b, right = 6+c+d ⇒3b=6+c+d(eq 3)

  • From eq 1 and eq 3

3(2d)=6+c+d

6d=6+c+d

5d=6+c (eq 4)

  • From eq 2 and eq 3

3(2+c)=6+c+d

6+3c=6+c+d

2c=d (eq 5)

  • From eq 4 and eq 5

5(2c)=6+c

10c=6+c

9c=6

c = 2/3

  • input eq 5

d = 2 x 2/3

d = 4/3

  • input eq 1

b = 2 x 4/3

b = 8/3

The equation

aZn + bHNO₃⇒ Zn(NO₃)₂ + cNO + dH₂O to

Zn + 8/3HNO₃⇒ Zn(NO₃)₂ + 2/3NO + 4/3H₂O x 3

3Zn + 8HNO₃⇒ 3Zn(NO₃)₂ + 2NO + 4H₂O

8 0
3 years ago
Determine the hydrogen ion concentration of solution B. [1]
bazaltina [42]
The hydrogen Ion concentration of solution  B is 
1.0 x  10^-5 or 0.000 010 M

You can see that this will be proportional to the amount of B's PH compared to A's

hope this helps
8 0
3 years ago
A student dissolves equal amounts of salt in equal amounts of warm water, room-temperature water, and ice water. Which result is
Tanya [424]

doesnt salt desolve ice? so wouldn't the salt dissolve in the ice water?

6 0
3 years ago
Read 2 more answers
A gas mixture with 4 mol of Ar, x moles of Ne, and y moles
maks197457 [2]

Answer:

a) \Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

b) x=4

c) \Delta G_{max}=-2721.9 J/mol

Explanation:

Gas mixture:

n_{Ar}= 4 mol

n_{Ne}= x mol

n_{Xe}= y mol

n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}

n_{Ne} + n_{Xe}=2*n_{Ar}

x + y=8 mol

y=8 mol- x

Mol fractions:

x_{Ar}=\frac{4 mol}{12 mol}=1/3

x_{Ne}=\frac{x mol}{12 mol}=x/12

x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12

Expression of \Delta G_{mixing}

\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)

\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]

\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

Expression of \Delta G_{max}

\frac{d \Delta G_{mixing}}{dx}=0

\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]

0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)

0=[ln (x/12)-ln ((8-x)/12)

ln (x/12)=ln ((8-x)/12)

x=(8-x)

x=4

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]

\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]

\Delta G_{max}=-2721.9 J/mol

4 0
3 years ago
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