Need help can somebody help me please

the problem say what is the volume (in kL) of 3.7505 x 10^4 mg of iron? Iron has a density of 7.87 g/ml. Use correct significant

Elis [28]

So first find the volume

so 7.78g/ml and (3.7505 times 10^4) grams
therefor we dividde (3.7505 times 10^4) by 7.78 and get how many ml

the answer is 4820.69 ml

kL means kilo lieters
kilo=1000
kL=1000Liters
ml=milileters=1/1000 leiter
1000ml=1L
therfor
1kL=1000L
1000L=1000 times (1000ml)
1000L=1,000,000ml
1kL=1,000,000ml

so to convert to kL divide 4820.69 by 1,000,000
0.00482069
convert to scientifiic notation
4.8 times 10^-3

the answer is 4.8 times 10^-3 kL

4 0

2 years ago

A covalent bond is formed by the sharing of _______ between atoms, whereas an ionic bond is formed by the _______.

OleMash [197]

Electrons and electrons

7 0

3 years ago

Copper sulfate can crystalize as an anhydrate (no water in the crystal) or as three different hydrates (varying amounts of water

Maurinko [17]

Answer:

Two factors that might have a affect of which copper sulphate mineral will occur at a given location is:

A. Copper sulphate high solubility in water

B. Also it binds nicely with the sediments or the crystal.

Explanation:

As it is mentioned here that copper sulphate can be crystallized as an anhydrate which means that their is no waterin those crystals or can be as of those three different hydrates whose crystal structure varies with the amount of water present in it.

The four forms are also given of the copper sulphate are:

Bonatite
Boothite
Chalcanthite
Chalcocyanite

So, the two factors that might give an affect which type of copper sulphate mineral willoccur at a given location is:

A. The copper sulphate high solubility in water.

B. It binds extremely nicely with the sediments or say to the crystal. It is also regulated by plants.

8 0

2 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$