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Y_Kistochka [10]
4 years ago
10

the general name for a substance added to a reaction that affects the rate but is not consumed in the reaction is called a_____.

complex constituent reactant catalyst
Chemistry
2 answers:
stiks02 [169]4 years ago
8 0

It's complex, google :)

cluponka [151]4 years ago
4 0

Answer:

complex

Explanation:

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Answer the four questions to figure out the four digit code
denis-greek [22]

Answer:

question 1 =c

question 2 =a

question 3 =d

question 4 =b

8 0
1 year ago
How much time does it take for a truck to slow down if its acceleration is -15m/s2 and the change in velocity is -30m/s
damaskus [11]

Answer:

2s

Explanation:

Given parameters:

Acceleration of the truck = 15m/s²

change in Velocity  = 30m/s

Unknown:

Time taken for the body to slow down = ?

Solution:

Acceleration is the rate of change of velocity ;

  Acceleration  = \frac{change in velocity}{time}

    time  = \frac{change in velocity }{acceleration}  

Now insert the parameters and solve;

   time  = \frac{30}{15}  = 2s

5 0
3 years ago
A system gains 15.0 J of thermal energy and does 65.0 J of work on its surroundings.
Ivanshal [37]

Answer:

-50J

Explanation:

Q(in) - W(out) = ΔU

15 - 65 = -50

5 0
3 years ago
Which type of chemical bonding involves a transfer of electrons?
BigorU [14]

Answer:

ionic bond

Explanation:

7 0
2 years ago
Read 2 more answers
Suppose that 10.0 mol C2H6(g) is confined to 4.860 dm3 at 27 °C. Predict the pressure exerted by the ethane from (i) the perfect
Afina-wow [57]

Answer:

P=35.16

Z=4.6

Explanation:

Hello,

In this case, since the VdW equation is:

P=\frac{nRT}{V-n*b}-a(\frac{n}{V} )^2

Since the moles are 10.0 moles, the temperature in K is 300.15 K and the volume is liters is also 4.860 L (1 dm³= 1L), the pressure exerted by the ethane is:

P=\frac{10.0mol*0.082\frac{atm*L}{mol*K}*300.15K}{4.860mol-10.0mol*0.0651\frac{L}{mol} }-5.507\frac{atm*L^2}{mol^2}(\frac{10.0mol}{4.86L} )^2\\\\P=58.48atm-23.3atm\\\\P=35.16

Thus the compression factor turns out:

Z=\frac{PV}{RT}=\frac{23.3atm*4.86L}{  0.082\frac{atm*L}{mol*K}*300.15K}\\\\Z=4.6

Regards.

8 0
3 years ago
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