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Tema [17]
3 years ago
10

In order from LEAST density to most in order

Chemistry
1 answer:
baherus [9]3 years ago
4 0
Ping pong ball - 0.0840 g
Bottle cap - 0.92 g
Marble - 2.711 g
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Calculate the mass of 3.4 moles of nitric acid (hno3). explain the process or show your work by including all values used to det
balu736 [363]

Answer:

6.34917360^25g

Explanation:

It's been a while since I've done this type of problem so I'm not making any promises that its right hahaha, but I hope it helps anyway. Please let me know whether I'm right or not!

6 0
1 year ago
Sodium carbonate (NaCO3) is sometimes used as a water-softening agent. Suppose that a worker prepares a 0.730 M solution of NaCO
sergey [27]

The molarity of a solution is the number of moles of a substance divided by the volume in liters prepared.

molarity=\frac{n}{V}, where n is number of moles and V is the volume in liters.

In order to calculate the mass of solute we need to convert the volume and molarity to moles

1.421 L solution \times\frac{0.0730 moles}{1 Lsolution}= 1.037 mol NaCO_3

Now that we have moles we use the relative formula mass of NaCO₃, We have 1 Na atom, 1 C atom and 3 O atoms, thus

M_r= (1\times 22.99) + (1\times 12.00) + (3\times 16.00)= 82.99g/mol

1.037 \times\frac{82.99g}{mol} = 86.1g

5 0
3 years ago
Read 2 more answers
Determine whether a gas sample was composed of hydrogen or oxygen.
Nuetrik [128]
The test for this is fairly simple.
We take a glowing match or splint near the gas sample, if the glow intensifies, oxygen is present. 
If a lit splint or match goes out with a popping sound, this means that hydrogen is present.
5 0
3 years ago
18. Which metal is capable of forming more than one cation? <br> Li<br> Ba<br> Al <br> Sn
maw [93]

Answer:

Li

Explanation:

3 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
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