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Liula [17]
3 years ago
13

WILL GIVE BRAINLIEST

Chemistry
1 answer:
natima [27]3 years ago
7 0

Answer:

1. True

2. resting position

3. 5 cm

4. Compressions

5. false

6. 7 m

7. true

8. second one has more amplitude

9. True

10. true

Explanation:

You might be interested in
An aqueous solution containing 17.5 g of an unknown molecular compound in 100.g water was found to have a freezing point of-1.8°
Zigmanuir [339]

Answer:

The answer to your question is 178.6 g

Explanation:

Data

ΔT = 1.8 °C

mass = 17.5 g

mass of water = 100 g

Kc = 1.86

Process

1.- Calculate the molality using the following formula

    ΔTc = mKc

solve for m

    m = ΔTc/Kc

substitution

     m = 1.8/1.86

result

     m = 0.968

2.- Calculate the number of moles

    m = # of moles/kg of solvent

kg of solvent = 0.1 kg

     # of moles = m x kg of solvent

     # of moles = 0.968 x 0.1

     # of moles = 0.0968

3.- Calculate the molar mass

        x g molar mass --------------------- 1 mol

       17.5 g                  --------------------- 0.0968 moles

       x = (1 x 17.5)/0.0968

      x = 178.6 g

             

4 0
3 years ago
Which of the following changes would have no effect on the equilibrium position of the reaction below? 2 NOBR (g) 2 NO (g)+ Br2
nasty-shy [4]

Answer:

D) cutting the concentrations of both NOBr and NO in half

Explanation:

The equilibrium reaction given in the question is as follows -

2NOBr ( g ) ↔  2NO ( g ) + Br₂ ( g )

The equilibrium constant for the above reaction can be written as -

K = [ NO ]² [ Br₂ ] / [ NOBr ] ²

Therefore from the condition given in the question , the changes that will not affect the equilibrium will be , reducing the concentration of both NOBr and NO to half ,

Hence ,

the new concentrations are as follows -

[ NoBr ] ' = 1/2 [ NoBr ]

[ NO ] ' = 1/2  [ NO ]

Hence the new equilibrium constant equation can be written as -

K ' = [ NO ] ' ² [ Br₂ ] / [ NOBr ]  ' ²

Substituting the new concentration terms ,

K ' = 1/2  [ NO ] ² [ Br₂ ] / 1/2 [ NoBr ]  ²

K ' = 1/4  [ NO ] ² [ Br₂ ] / 1/4 [ NoBr ]  ²

The value of 1 / 4 in the numerator and the denominator is cancelled -

K ' =   [ NO ] ² [ Br₂ ] /  [ NoBr ]  ²

Hence ,

K' = K

5 0
4 years ago
Bob rides his horse 52 km in 3 hours 15 minutes. What is his average speed in kilometers per hour?​
laila [671]
I think the answer is 16.51
7 0
3 years ago
Several balloons are inflated with helium to a volume of 0.75 L at 27°C. One of the balloons was found several
defon

Answer:

V₂ = 0.74 L

Explanation:

Given data:

Initial volume of balloon = 0.75 L

Initial temperature of balloon = 27°C (27+273.15 K = 300.15 K)

Final volume of balloon = ?

Final temperature = 22°C (22+273.15 K = 295.15 k)

Solution:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 0.75 L × 295.15 k / 300.15 K

V₂ = 221.36 L.K / 300.15 K

V₂ = 0.74 L

The given problem is solved through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

7 0
3 years ago
Determine the mass of grams of led(II) sulfate that will dissolve in 2.50 x 10^2 ml water. Ksp= 1.8 x 10^-8
ki77a [65]

Answer:

9.86*10^(-3) g

Explanation:

PbSO4 ----> Pb^(2+) + SO4^(2-)

                     s               s

Ksp = s²

s =√Ksp = √(1.8*10^-8) = 1.3*10^(-4) mol/L

The molar solubility PbSO4 = 1.3*10^(-4) mol/L.

2.50 *10^2 mL *1L/10³mL =0.250L

1.3*10^(-4)mol/L *0.250L*303.3 g/mol = 9.86*10^(-3) g

7 0
3 years ago
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