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White raven [17]
3 years ago
15

1. Balance the reaction: Ag + H2S + O2 Ag2S + H2O

Chemistry
1 answer:
Murrr4er [49]3 years ago
8 0
It’s 0 because if u add them and divide 0 it’s 0
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Which of the following is true if the net force on an object is zero? (1 pt) a. The object must be at rest. b. The object’s spee
prohojiy [21]

Explanation:

the object must be at rest

5 0
2 years ago
Na +H₂O- NaOH +H₂ balance
Whitepunk [10]

Hey there!

Na + H₂O → NaOH + H₂

First, balance O.

One on the left, one on the right. Already balanced.

Next, balance H.

Two on the left, three on the right. Let's add a coefficient of 2 in front of NaOH and a coefficient of 2 in front of H₂O, so we have 4 on each side.

Na + 2H₂O → 2NaOH + H₂

Lastly, balance Na.

One on the left, two on the right. Add a coefficient of 2 in front of Na.

2Na + 2H₂O → 2NaOH + H₂  

This is our final balanced equation.

Hope this helps!

7 0
3 years ago
A liquid-phase isomerization is carried out in a 1000-gal CSTR that has a single impeller located halfway down the reactor. The
igomit [66]

Answer:

The conversion achieved for the first CSTR impeller is 0.382

Discrepancy = 0.188

Explanation:

The impeller divides the CSTR into 2 equal reactors of volume 500gal

Using V = FaoX/ (-ra)

500gal = Fao×Xa/[(KCao^2( 1 -X1)^2]

500gal = CaoVoX1/ KCao^2(1-X1)

500gal= 500gal × X1'/(1 - X1)^2

(1 -X1)^2 = X1

X1^2 - 3X1 + 1 = 0

X1= 0.382

Conversion achieved in the first CSTR is 0.382

Actual measured CSTR = 57% =57/100=0.57

Discrepancy in the conversions= 0.57 -0.383 =0.188

8 0
2 years ago
Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm
8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

\frac{dV}{dt}=10\frac{cm^3}{s}

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

We proceed to its differentiation:

\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}

Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

Best regards.

4 0
3 years ago
5,6-dimethyl-2-heptyne <br><br> Due in 1 hr pls help me
mina [271]

Answer:

Please see the attached pictures.

Explanation:

☆ To ensure that each carbon has 4 bonds, fill the other bonds with Hs.

4 0
2 years ago
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