Which is the formula for nitrogen trihydride? NH3 N3H 3NH N3H3

Ow many coulombs of charge are needed to produce 35.2 mol of solid zinc?

Natasha2012 [34]

Answer: Zn(2+) + 2e(-) -------> Zn 1 mole of Zn is deposited by 2F of electricity ... so 48.9 mole of Zn will be deposited by 48.9 X 2F = 97.8 F of electricity... as 1F = 96500 C so 97.8 F = 97.8 X 96500 = 9437700 C of electricity...

8 0

3 years ago

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If the solubility of sugar at 25ºC is 211 g/100 g H2O, what mass of sugar can be dissolved in 300 g of H2O?

ladessa [460]

Answer:

633 grams of sugar can be dissolved in 300 g of H₂O

Explanation:

Solubility is the measure of the ability of a certain substance to dissolve in another and form a homogeneous system. Solubility is then the maximum amount of a solute that a solvent can receive and is expressed by concentration units.

The rule of three or is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them. That is, what is intended with it is to find the fourth term of a proportion knowing the other three. Remember that proportionality is a constant relationship or ratio between different magnitudes.

If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied. To solve a direct rule of three, the following formula must be followed:

a ⇒ b

c ⇒ x

Then:

$x=\frac{c*b}{a}$

You can apply the rule of three as follows: if by definition of solubility in 100 grams of H₂O there are 211 grams of sugar, in 300 g of H₂O how much sugar is there?

$sugar=\frac{300 grams of H_{2}O *211 grams of sugar}{100 grams of H_{2}O}$

sugar= 633 grams

633 grams of sugar can be dissolved in 300 g of H₂O

7 0

2 years ago

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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

How are the terms molar mass and atomic mass different from one another

Ronch [10]

1.Molar mass is the mass of one mole per single element while atomic mass is the mass of an atom at rest or is the number of protons and neutrons.

2.Molar mass is measured in grams per mole while atomic mass is “unitless.”

3.Atomic mass is measured via mass spectrometry while molar mass is computed via atomic weight.

8 0

3 years ago

Which tool should Darla and Rob use to measure the force that pulls a car down the map

svetoff [14.1K]

Answer:

C. A Spring Scale

Explanation:

Using process of elimination, we can quickly decide that a stopwatch and a ruler will not be useful in measuring the force. This leaves us with either the spring scale or a balance scale. A balance scale is used to compare two weights, so this is eliminated. That leaves us with a spring scale. This is because we can attached the spring scale to the car and when we let it go, we can record the force. I hope this helps!

5 0

2 years ago