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3 years ago

A container has the dimensions of 10cm x 80mm x 0.15m. The density of

Chemistry

1 answer:

vesna_86 [32]3 years ago

8 0

Answer:

2.52kg

Explanation:

80mm = 8 cm

0.15m = 15 cm

volume of the container = 10m × 8cm × 15cm

= 1200 cubic centimetres

mass = density × volume

= 2.1 × 1200

= 2520g

2520/ 1000

= 2.52kilograms

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Water travels through many different pathways within the hydrologic cycle. For example, water vapor in the atmosphere could cond

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It would most likely be D because that’s how most of the things would be processed but sorry if I’m wrong.

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2 years ago

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Calculate the pH for each of the cases in the titration of 25.0 mL of 0.180 M pyridine, C 5 H 5 N ( aq ) with 0.180 M HBr ( aq )

Oxana [17]

Answer:

Explanations

Calculate the pH for each of the following cases in thetitration of 25.0 mL of 0.100 M pyridine, C5H5N(aq) with 0.100 MHBr(aq):

a.Before and addition of HBr

b.After addition of 12.5ml HBr

c.After addition of 15ml HBr

d.After addition of 25ml HBr

e.After addition of 33ml HBr

SOLUTION ;;;

Kb of pyridine =1.5*10^-9

a)let the dissociation be x.so,

kb=x^2/(0.1-x)

or 1.5*10^-9=x^2/(0.1-x)

or x=1.225*10^-5

so,

[OH-]=1.225*10^-5

so,

pOH=-log([OH-])

so pH=9.088

b)now this will effectively behave as a buffer

pKb=8.82

so pOH=pKb+log(salt/acid)

=8.82+log((12.5*0.1)/(25*0.1-12.5*0.1))

=8.82

so pH=14-pOH

=5.18

c)again using the same equation as the above,

pOH=pKb+log(salt/acid)

=8.82+log((15*0.1)/(25*0.1-15*0.1))

so pH=14-9

d)now the base is completely neutralised.so,

concentration of the salt formed=0.1/2

=0.05 M

so,

pH=7-0.5pKb-0.5log(C)

=7-0.5*8.82-0.5*log(0.05)

=3.24

e)concentration of H+=(33*0.1-25*0.1)/(33+25)

=0.01379

so pH=-log(0.01379)

=1.86

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3 years ago

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Consider the following generic reaction for which Kp = 5.51 × 105 at 25°C:

lora16 [44]

Answer:

$K_c=1.35\times 10^7$

Explanation:

The relation between Kp and Kc is given below:

$K_p= K_c\times (RT)^{\Delta n}$

Where,

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

$2R_{(g)}+A_{(g)}\rightleftharpoons2Z_{(g)}$

Given: Kp = $5.51\times 10^5$

Temperature = 25°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

T = (25 + 273.15) K = 298.15 K

R = 0.082057 L atm.mol⁻¹K⁻¹

Δn = (2)-(2+1) = -1

Thus, Kc is:

$5.51\times 10^5= K_c\times (0.082057\times 299)^{-1}$

$K_c\frac{1}{24.535043}=551000$

$K_c=13518808.69=1.35\times 10^7$

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3 years ago

Explain and give examples of theories and laws (big bang molecular clock etc.)

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3. Ohms law- says the current passing through an area is directly proportional to the voltage

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3 years ago

_KOH+_H3PO4=_K3PO4+_H2O

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The balanced chemical equation for given reaction is ~

$\sf{3KOH + H_3PO_4 \rightarrow K_3PO_4 + 3H_2O}$

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2 years ago