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3 years ago

Why could a diet low in carbohydrates be dangerous?

1 answer:

4 0

Yo!
Im not sure 'bout your answer but 60 percent the answer can be,
The body could take in too much dietary fiber, which harms the cell.

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A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.

Lubov Fominskaja [6]

Answer:

$Ag_2SO_4$

Explanation:

Formula for the calculation of no. of Mol is as follows:

$mol=\frac{mass\ (g)}{molecular\ mass}$

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

$mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol$

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

$mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol$

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

$mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol$

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

$Ag, \frac{0.05305}{0.02657} \approx 2$

$S, \frac{0.02657}{0.02657} \approx 1$

$O, \frac{0.10594}{0.02657} \approx 4$

Therefore, empirical formula of the compound = $Ag_2SO_4$

7 0

3 years ago

Read 2 more answers

If 100% is not passed from one trophic level to the next, where does this energy go?

Sonja [21]

D is the answer to question

6 0

3 years ago

Read 2 more answers

Can you please help me

rosijanka [135]

Shure what you need help with

7 0

3 years ago

Notice that " s o 4 " appears in two different places in this chemical equation. s o 2− 4 is a polyatomic ion called "sulfate."

Amanda [17]

A "3" should but put in front of
"cas o 4 "

6 0

3 years ago

From these two reactions at 298 K, V2O3(s) + 3CO(g) → 2V(s) + 3CO2(g); ΔH° = 369.8 kJ; ΔS° = 8.3 J/K V2O5(s) + 2CO(g) → V2O3(s)

gregori [183]

Answer:

ΔG° = -133,1 kJ

Explanation:

For the reactions:

(1) V₂O₃(s) + 3CO(g) → 2V(s) + 3CO₂(g); ΔH° = 369,8 kJ; ΔS° = 8,3 J/K

(2) V₂O₅(s) + 2CO(g) → V₂O₃(s) + 2CO₂(g); ΔH° = –234,2 kJ; ΔS° = 0,2 J/K

By Hess's law it is possible to obtain the ΔH° and ΔS° of:

2V(s) + 5CO₂(g) → V₂O₅(s) + 5CO(g)

Substracting -(1)-(2), that means:

ΔH° = -369,8 kJ - (-234,2 kJ) = -135,6 kJ

ΔS° = - 8,3 J/K - 0,2 J/K = -8,5 J/K

Using: ΔG° = ΔH° - TΔS° at 298K

ΔG° = -135,6 kJ - 298K×-8,5x10⁻³kJ/K

ΔG° = -133,1 kJ

I hope it helps!

7 0

3 years ago