All categories

2 years ago

Can someone please try and help me with this? I'll give brainliest if I can

Chemistry

1 answer:

sineoko [7]2 years ago

6 0

Answer:

boom Corvette Corvette pop in the popular vet like that like that why you walk like that why you talk like that like I said boom boom boom I'm in the cold Mira, she needs to be back I remember seeing everyone Nee u i

You might be interested in

What is the coordination number for each of the following complexes or compounds? [Co(NH3)4(H2O)2]3+ [Cr(EDTA)]− [Pt(NH3)4]2+ Na

Burka [1]

Answer:

[Co(NH₃)₄(H₂O)₂]³⁺: coordination number = 6.

[Cr(EDTA)]⁻: coordination number = 6.

[Pt(NH₃)₄]²⁺: coordination number = 4.

Na[Au(Cl)₂]: coordination number = 2.

Explanation:

[Co(NH₃)₄(H₂O)₂]³⁺:

In this complex, Co is bonded with 4 molecules of NH₃ (with 4 coordinate bonds, one bond for each molecule) and 2 molecules of H₂O (with 2 coordinate bonds, one bond for each molecule) forming the complex with 6 coordinate bonds.

∴ coordination number = 6.

[Cr(EDTA)]⁻:

In this complex, Cr is bonded with 1 molecules of EDTA (with 6 coordinate bonds, 4 O atoms and 2 N atoms in EDTA molecule).

∴ coordination number = 6.

[Pt(NH₃)₄]²⁺:

In this complex, Pt is bonded with 4 molecules of NH₃ (with 4 coordinate bonds, one bond for each molecule).

coordination number = 4.

Na[Au(Cl)₂]:

In this complex, Au is bonded with 2 atoms of Cl (with 2 coordinate bonds, one bond for each atom).

coordination number = 2.

3 0

2 years ago

A 46.9 gram sample of a substance has a volume of about 3.5 centimeters3. It is solid at a room temperature of 23ºC. Out of the

horrorfan [7]

Answer : (C) Hafnium is the most likely identity of the given substance.

Solution : Given,

Mass of given substance (m) = 46.9 g

Volume of given substance (V) = 3.5 $Cm^{3}$

First, find the Density of given substance.

Formula used :

$Density=\frac{\text{Mass of given substance}}{\text{Voume of given substance}}$

Now,put all the values in this formula, we get

$Density=\frac{46.9 g}{3.5 Cm^{3} }$ = 13.4 g/ $Cm^{3}$

So, we conclude that the density of given substance (13.4 g/ $Cm^{3}$ ) is approximately equal to the density of Mercury and Hafnium (13.53 and 13.31 g/ $Cm^{3}$ respectively).

According to the question the substance is solid at room temperature but Mercury is liquid at room temperature. So, Mercury is not identical to the given substance.

Another element i.e, Hafnium is the element whose density is approximately equal to the given substance and also solid at room temperature. And we know that the melting point of solid is high.

So, Hafnium is the most likely element which is the identity of the given substance.

3 0

2 years ago

Read 2 more answers

How many atoms are in 1.00 molecules of he

irina1246 [14]

Answer:

Explanation:

6 0

3 years ago

Read 2 more answers

1.While cooking, if the bottom of the vessel is getting blackened on the outside.