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3 years ago

What is the H30* concentration of 0.027 M Ca(OH)2?

Chemistry

2 answers:

Georgia [21]3 years ago

8 0

Answer:

1.9 x 10⁻¹³M

Explanation:

Given 0.027M Ca(OH)₂(aq) => 0.027M Ca⁺²(aq) + 2(0.027M) OH⁻(aq)

2(0.027M) OH⁻(aq) = 0.054M OH⁻(aq).

from [H⁺][OH⁻] = 1 x 10⁻¹⁴ at STP Conditions (0°C, 1Atm)

∴{H⁺] = [H₃O⁺] = (1 x 10⁻¹⁴/0.054)M = 1.85 x 10⁻¹³M or 1.9 x 10⁻¹³at 0°C & 1Atm pressure.

puteri [66]3 years ago

6 0

The correct answer is C. 1.9 x 10-13 M
• I just summed down the answer from the other guy in case people didn’t get the answer

Have a nice day :)

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Among timber, wind, and oil, which is a fossil fuel and which is not?

Cloud [144]

Answer:

Oil is a fossil fuel and timber or wood and wind is not. Hope this helps!

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2 years ago

Read 2 more answers

FIND THE RATIO BY MASS OF THE COMBINING ELEMENTS IN THE FOLLOWING COMPOUNDS

boyakko [2]

The given compound is Aluminum sulfate, Al2(SO4)3:

Molar masses:

Aluminum = 27 g/mol
Sulfur = 32 g/mol
Oxygen = 16 g/mol

The total molar mass is 342 g/mol
The ratio by mass of the elements:

Aluminum = 27*2/342
= 0.16
Sulfur = (32*3)/342
= 0.28
Oxygen = (16*12)/342
= 0.56
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2 years ago

1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.

andriy [413]

Answer:

A. $m_{NH_3}^{theo} =1.50gNH_3$

B. $Y=82.2\%$

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

$N_2+3H_2\rightarrow 2NH_3$

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

$n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\ n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3$

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

$m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3$

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

$Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%$

Best regards!

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 $lb/ft^{2}$

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3 years ago

kotegsom [21]

Answer:

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Explanation:

yes it is true

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3 years ago