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pav-90 [236]
3 years ago
6

A wooden plank has a mass of 100kg force of 20 n was used to move it what is its acceleration​

Chemistry
1 answer:
krok68 [10]3 years ago
4 0

Answer:

a= 0.2m/s²

Explanation:

Fnet = ma

20 = 100a (divide both sides by 100)

a = 0.2m/s²

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3 years ago
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DENIUS [597]

Answer:

b. 760 g

Explanation:

The mass of the solution = 800 g

5% of NaCl by mass of the solution can be determined as follows;

    5% of 800  =  \frac{5}{100} × 800

                       = 5 × 8

                       = 40 g

The mass of NaCl in the solution is 40 g.

The mass of water = mass of solution - mass of NaCl

                               = 800 - 40

                              = 760 g

Therefore, the mass of water required is 760 g.

5 0
4 years ago
Calculate the standard entropy of vaporization of ethanol at its boiling point 285 K. The standard molar enthalpy of vaporizatio
Ivanshal [37]

Answer:

standard entropy of vaporization of ethanol = 142.105 J/K-mol

Explanation:

given data

enthalpy of vaporization of ethanol = 40.5 kJ/mol = 40.5 × 10^{3} J/mol

entropy of vaporization of ethanol boiling point = 285 K

to find out

standard entropy of vaporization of ethanol

solution

we get here standard entropy of vaporization of ethanol that is expess as

standard entropy of vaporization of ethanol ΔS = \frac{\Delta H}{T} .............1

here ΔH is enthalpy of vaporization of ethanol and  T is temperature

put value in equation 1

standard entropy of vaporization of ethanol ΔS =  \frac{40.5*10^3}{285}

standard entropy of vaporization of ethanol = 142.105 J/K-mol

3 0
4 years ago
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Sports drinks contain sugar, salt, and flavorings dissolved in water . They can be made more or less sweet by adding different a
givi [52]

Answer:

It's a mixture.

Explanation:

Answer from Edgenuity

8 0
3 years ago
When adjusted for any changes in δh and δs with temperature, the standard free energy change δg∘t at 2400 k is equal to 1.22×105
rosijanka [135]
When adjusted for any changes in δh and δs with temperature, the standard free energy change δg∘t at 2400 k is equal to 1.22×105j/mol, then the equilibrium constant at 2400 k is 2.21×10−3. The answer to the statement is 2.21×10−3.
6 0
3 years ago
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