What is the name of GaF3?

I really need help, I don’t know what to do on this

Shtirlitz [24]

Answer:

A: dependent variable

B: experimental group

C: independent variable

D: control variables

E: control group

Explanation:

7 0

3 years ago

A 100. 0 ml sample of 0. 10 m nh3 is titrated with 0. 10 m hno3. Determine the ph of the solution after the addition of 100. 0 m

DerKrebs [107]

The pH of the solution after the addition of 100. 0 ml of HNO3. The kb of NH3 is 1. 8 × 10-5. So, pH is 10.9 basic .

This neutralization occurs as the acid is added to the base:

NH3(aq)+ HNO3(aq)= NH4NO3(aq)+ H2O(l)

The initial moles of NH3present is given by,

nNH3= c×v= 0.10× 100/1000= 0.01m

The number of moles of

HNO3 added is given by:

nHNO3= c×v= 0.10× 100/1000= 0.001

It is clear from the equation that the acid and base react in a 1:1 molar ratio. So, the no. moles of NH3remaining will be 0.01 - 0.001= 0.009

The total volume is now

100.0+100.0= 200.0x cm3

The concentration of NH3 is given by,

[NH3]= c/v= 0.001/200/1000= 0.05x mol/l .

pOH= 12 (pKb− logb)

where b is the base's concentration.

Given that there is little dissociation, we can roughly compare this to the starting concentration.

pKb= −logKb= −log(1.8×10−5)= 4.744

pOH= 3.056

At 25∘xC, we know that, pH+ pOH=14

pH= 14− 3.056= 10.9

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8 0

1 year ago

What is the ph of a 0.25 m solution of c6h5nh2 given that its kb is 1.8 x 10-6?

IrinaK [193]

The pH a 0.25 m solution of C₆H₅NH₂ is equal to 3.13.

<h3>How do we calculate pH of weak base?</h3>

pH of the weak base will be calculate by using the Henderson Hasselbalch equation as:

pH = pKb + log([HB⁺]/[B])

pKb = -log(1.8×10⁻⁶) = 5.7

Chemical reaction for C₆H₅NH₂ is:

C₆H₅NH₂ + H₂O → C₆H₅NH₃⁺ + OH⁻

Initial: 0.25 0 0

Change: -x x x

Equilibrium: 0.25-x x x

Base dissociation constant will be calculated as:
Kb = [C₆H₅NH₃⁺][OH⁻] / [C₆H₅NH₂]

Kb = x² / 0.25 - x

x is very small as compared to 0.25, so we neglect x from that term and by putting value of Kb, then the equation becomes:

1.8×10⁻⁶ = x² / 0.25

x² = (1.8×10⁻⁶)(0.25)

x = 0.67×10⁻³ M = [C₆H₅NH₃⁺]

On putting all these values on the above equation of pH, we get

pH = 5.7 + log(0.67×10⁻³/0.25)

pH = 3.13

Hence pH of the solution is 3.13.

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5 0

2 years ago

CaBr + KOH – Ca(OH), + KBr (balance first) What mass, in grams, of

neonofarm [45]

Answer:

129.73 g of CaBr₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

CaBr₂ + 2KOH –> Ca(OH)₂ + 2KBr

Next, we shall determine the mass of CaBr₂ that reacted and the mass of Ca(OH)₂ produced from the balanced equation. This can be obtained as follow:

Molar mass of CaBr₂ = 40 + (80×2)

= 40 + 160

= 200 g/mol

Mass of CaBr₂ from the balanced equation = 1 × 200 = 200 g

Molar mass of Ca(OH)₂ = 40 + 2(16 + 1)

= 40 + 2(17)

= 40 + 34

= 74 g/mol

Mass of Ca(OH)₂ from the balanced equation = 1 × 74 = 74 g

SUMMARY :

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Finally, we shall determine the mass of CaBr₂ that react when 48 g of Ca(OH)₂ were produced. This can be obtained as follow:

From the balanced equation above,

200 g of CaBr₂ reacted to produce 74 g of Ca(OH)₂.

Therefore, Xg of CaBr₂ will react to produce 48 g of Ca(OH)₂ i.e

Xg of CaBr₂ = (200 × 48)/74

Xg of CaBr₂ = 129.73 g

Thus, 129.73 g of CaBr₂ were consumed.

6 0

3 years ago

I don't really understand this worksheet question.

Citrus2011 [14]

25/2 and 96/X
CROSS MULTIPLY.

2x=2,400.
divide by 2.
x=1,200.

you take the GIVEN MASS of an element, and you put it on top, the coefficient is what it’s over. i believe this is right

3 0

3 years ago