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Mars2501 [29]
3 years ago
10

Why should the wearing of uniforms be discontinued in highscho​

Chemistry
1 answer:
Sav [38]3 years ago
6 0
Because there is no need for them
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What is used in chemistry/science to heat or warm something?
DENIUS [597]
Several pieces of equipment can do this, including the Bunsen burner, laboratory oven, hot plate and incubator.
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3 years ago
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Select the correct pronoun.
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It is who! :D
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1. A car has a mass of 4000 kg. What force is required to make the car accelerate at 20 mss
shtirl [24]

Answer:

As Per Given Information

Mass of Car 4000 Kg

Acceleration of car 20m/s²

we have to find the force required to make the car car accelerate at given magnitude .

Using Formulae

  • Force = Mass × Acceleration

On Putting the value we obtain

↝ Force = 4000 × 20

↝ Force = 80000 N

↝ Force = 80KN

So, the force required to maintain the given acceleration is 80 KiloNewton .

8 0
3 years ago
Calculate the value of ΔG∘rxnΔGrxn∘ for the following reaction at 296 K. Ka = 2.9 × 10–8 and assume Ka does not change significa
oee [108]

Answer:

\Delta G_{rxn}=42.7\frac{kJ}{mol}

Explanation:

In this case, for the dissociation of hypochlorous acid, we know that the acid dissociation constant (Ka) is 2.9x10⁻⁸, which is related with the Gibbs free energy as shown below:

\Delta G_{rxn}=-RTln(K)

But in this case K is just Ka, therefore, at 296 K, it turns out:

\Delta G_{rxn}=-8.314\frac{J}{mol*K}*296K*ln(2.9x10^{-8})\\\\\Delta G_{rxn}=42.7\frac{kJ}{mol}

Such result, means that the reaction is nonspontaneous at the given temperature, it means it is not favorable (not easily occurring).

Best regards.

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3 years ago
What are the molality and mole fraction of solute in a 22.3 percent by mass aqueous solution of formic acid (HCOOH)?
Vanyuwa [196]

Answer:

Mole fraction for solute = 0.1, or 10%

Molality = 6.24 mol/kg

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22.3% by mass → In 100 g of solution, we have  22.3 g of HCOOH

Mass of solution = 100 g

Mass of solute = 22.3 g

Mass of solvent = 100 g - 22.3g = 77.7 g

Let's convert the mass to moles

22.3 g . 1mol/ 46 g = 0.485 moles

77.7 g. 1mol / 18 g = 4.32 moles

Total moles = 4.32 moles + 0.485 moles = 4.805 moles

Xm for solute = 0.485 / 4.805 = 0.100 → 10%

Molality → mol/ kg → we convert the mass of solvent to kg

77.7 g.  1 kg / 1000g = 0.0777 kg

0.485 mol / 0.0777 kg = 6.24 m

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