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2 years ago

14

According to one acid base theory the water acts as ?

1 answer:

marshall27 [118]2 years ago

3 0

Answer By observing both concepts, water is acting as a base

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A theory is a well-established, data-backed idea that has been shown true many times but cannot be 100% proven

mafiozo [28]

Well if you were asking it was a true or false questions the answer is: True

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3 years ago

Which of these best describes the state of the substance represented by State Z?

romanna [79]

A) it is a solid because strong attractive forces prevent particles from moving.

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3 years ago

50 grams of acetic acid C2H4O2 are dissolved in 200 g of water. Calculate the weight % and mole fraction of the acetic acid in t

Len [333]

$50g$ $of$ $C_2H_4O_2$ $in$ $200g$ $H_2O$

so:

$\frac{50}{200+50}+100\%= \frac{50*100}{250}\%=20\%$

$MW of acid = 2*C+4*H+2*O = 2*12+4*1+2*16=$

$=24+4+32=60g/mol$

so:

$\frac{50g}{60g/mol}\approx0.83moles$

it means that

in $50g$ of acid there is $\approx0.83$ moles of acid

$MW of H_2O = 2*H+O=2*1+16=2+16=18g/mol$

so:

$\frac{200g}{18g/mol}\approx11.11moles$

it means that:

in $200g$ of water there is $\approx11.11$ moles of water

therefore:

$\frac{0.83mol}{11.11mol+0.83mol}= \frac{0.83mol}{11.94mol}=0.069$

So your answers are:

$20\%$

and the mole fraction is:

$0.069$

6 0

3 years ago

The mass of a length of metal wire is 45 g. When the wire is placed in a graduated

babunello [35]

5 ml=5 cm³ , density=mass/volume=45/5= 9 g/cm³ therefore answer B

7 0

4 years ago

Read 2 more answers

A 12.0 g sample of a metal is heated to 90.0 ◦C. It is then dropped into 25.0 g of water. The temperature of the water rises fro

deff fn [24]

Answer:

The specific heat of the metal is 0.34 J/g°C

Explanation:

Step 1: Data given

Mass of the metal = 12.0 grams

Mass of the water = 25.0 grams

Initial temperature of the metal = 90.0 °C

Initial temperature of the water = 22.5 °C

Final temperature = 25 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of the metal

Qlost = Qgained

Q = m*c*ΔT

Qmetal = -Qwater

m(metal) *c(metal)* ΔT(metal) = -m(water) * c(water) *ΔT(water)

⇒ with mass of metal = 12.0 grams

⇒ with c(metal) = TO BE DETERMINED

⇒ with ΔT(metal) = T2 - T1 = 25.0°C - 90.0 °C = -65.0 °C

⇒ with mass of water = 25.0 grams

⇒ with c(water) = 4.184 J/g°C

⇒ with ΔT(water) = T2 - T1 = 25.0 - 22.5 °C = 2.5 °C

12.0 * c(metal) * -65.0 °C = -25.0g * 4.184 J/g°C * 2.5°C

-780.0 * c(metal) = -2615 ( 2.6*10^3 with sig figs)

c(metal) = 0.335 (=0.34 with sig figs)

The specific heat of the metal is 0.34 J/g°C

3 0

3 years ago