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aleksley [76]
3 years ago
7

Made from other rock fragments​

Chemistry
1 answer:
zysi [14]3 years ago
6 0

Answer:

A rock fragment, in sedimentary geology, is a sand-sized particle or sand grain that is made up of multiple grains that are connected on the grain scale. These can include grains which are sand-sized themselves (a granitic rock fragment), or finer-grained materials (shale fragments). at least thats what i thought u were asking

Explanation:

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A substance made of both metal and nonmetal atoms ?
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A metalloid is a metal and a nonmetal
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1. A chemical bond between metals and nonmetals; valence electrons are transferred ____.
Law Incorporation [45]

Answer:

1. Ionic bonding

2. Covalent bonding

3. Metallic bonding

Explanation:

Ionic bonding also referred to as electrovalent bonding is a kind of chemical bonding that involves the transfer of electrons between the valence shells of two elements with a large electronegativity difference usually a metal and a nonmetal.

For example an ionic bonding scenario might play out between a group one metal and a group seven halogen. While group one metals have one electron hindering their stability, group seven halogens need that one electron that could make them achieve this stability. It is this that causes them to come together in a way where the electron is transferred completely from the valence shell of the group 1 atom and accepted into the valence shell of the group 7 halogen.

Covalent bonding involves the sharing of electrons between atoms of comparable electronegativities. The electro negativity difference is not large enough to permit the total movement of the electrons and hence the electrons are then controlled by the nuclei of the two atoms

Between two metals, what we have is called the metallic bonding

7 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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