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2 years ago

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Question 3

1 answer:

Oksanka [162]2 years ago

4 0

Answer:

maybe the third option?

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What is the volume of 0.250 m hydrochloric acid required to react completely with 20.0 ml of 0.250 m ca(oh)2?

Soloha48 [4]

The balanced equation for the above neutralisation reaction is as follows;
Ca(OH)₂ + 2HCl ----> CaCl₂ + 2H₂O
Stoichiometry of Ca(OH)₂ to HCl is 1:2
number of Ca(OH)₂ moles reacted - 0.250 mol/L x 20.0 x 10⁻³ L = 5.00 x 10⁻³ mol
according to molar ratio of 1:2
number of HCl moles required = 2 x number of Ca(OH)₂ moles reacted
number of HCl moles = 5.00 x 10⁻³ x 2 = 10.0 x 10⁻³ mol
molarity of HCl solution - 0.250 M
there are 0.250 mol in volume of 1 L
therefore 10.0 x 10⁻³ mol in - 10.0 x 10⁻³ mol / 0.250 mol/L = 40.0 mL
40.0 mL of 0.250 M HCl is required

5 0

3 years ago

Show work, thanks

GarryVolchara [31]

Answer:

Q.1

Given-

Volume of solution-1 L

Molarity of solution -6M

to find gms of AgNO3-?

Molarity = number of moles of solute/volume of solution in litre

number of moles of solute = 6×1= 6moles

one moles of AgNO3 weighs 169.87 g

so mass of 6 moles of AgNO3 = 169.87×6=1019.22

so you need 1019.22 g of AgNO3 to make 1.0 L of a 6.0 M solution

7 0

2 years ago

Find the momentum of 5 kg bowling ball at 10 m/s.

leva [86]

30 kg m/s

momentum = mass x velocity = 10 x 3 m/s =30 kg m/s

5 0

3 years ago

Nountains are formed at continental plates that converge because they have similar _______.

zheka24 [161]

Because of how they grow

5 0

3 years ago

Air at 7S°F and14.6 though a cfm(cubic ft per minute) and the The flow rate is 48000 psia flows though a rectangular duct of Ix2

IgorLugansk [536]

Explanation:

The given data is as follows.

Air is at $75^{o}F$ and 14.6 psia.

$\varepsilon$ = 0.00015 ft, Flow rate, (Q) = 48000 $ft^{3}/m$

(a) Formula to calculate hydraulic radius $(r_{H})$ is as follows.

$r_{H} = \frac{\text{free flow area}}{\text{wet perimeter}}$

= $\frac{2 \times 1}{2(1) + 2(2)}$

= $\frac{1}{3}$ ft

Formula for equivalent diameter is as follows.

$D_{eq} = 4 \times r_{H}$

= $4 \times \frac{1}{3} ft$

= $\frac{4}{3}$ ft

(b) Formula for velocity floe is as follows.

Q = VA

V = $\frac{Q}{A}$

= $\frac{48000}{2 \times 1}$ ft/min

= 24000 ft/min

(c) Formula to calculate Reynold's number is as follows.

$R_{e}$ = $\frac{D \times V \times \rho}{\mu}$

= $\frac{\frac{4}{3} \times 24000 \times 0.0744}{0.0443}$ (as $\rho = 0.0744 lb/ft^{3}$ and $\mu$ = 0.0443 lb/ft. hr)

= 53742.66 hr/min

As 1 hr = 10 min. So, $53742.66 hr/min \times \frac{60 min}{1 hr}$

= 3224559.6

(d) Formula to calculate pressure drop $(\Delta P)$ is as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

Putting the given values into the above formula as follows.

$\frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}$

= $\frac{4 \times 0.00015 \times 100 \times 0.0744 \times (24000)^{2}}{2 \times \frac{4}{3} \times {4}{3}}$

= 6.238 $lb/ft^{2}$

5 0

3 years ago