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3 years ago

Rust is soft and ---------

Chemistry

1 answer:

Lostsunrise [7]3 years ago

7 0

I think changes to blue in colour I hope it helps u

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C8H18+ O2> CO2+ H2O how to make this balanced?

Sonbull [250]

Product:
C: 1
O:3
H: 2

reactant:
C: 8
H: 18
O: 2

first you need to put probably an eight in front of the Carbon.
Next, add a coefficient of 9 to Hydrogen
finally, add a 3 INSTEAD of a 2 to Oxygen

should look like:
C8H18+O3>C8O2+H9O

7 0

3 years ago

When a __________ is completely dissolved in a __________, a solution is formed.

ValentinkaMS [17]

When a solute is completely dissolved in a solvent, a solution is formed.

8 0

3 years ago

What is the empirical formula for a compound that is 83.7% carbon and 16.3% hydrogen?

zubka84 [21]

Hello!

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100 g solution)

C: 83.7% = 83,7 g
H: 16.3% = 16.3 g

Let us use the above mentioned data (in g) and values will be converted to amount of substance (number of moles) by dividing by molecular mass (g / mol) each of the values, lets see:

$C: \dfrac{83.7\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} \approx 6.975\:mol$

$H: \dfrac{16.3\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 16.3\:mol$

We note that the values found above are not integers, so let's divide these values by the smallest of them, so that the proportion is not changed, let's see:

$C: \dfrac{6.975}{6.975} = 1$

$H: \dfrac{16.3}{6.975} \approx 2.3$

Note: So the ratio in the smallest whole numbers of carbon to hydrogen is 3:7, thus, the minimum or empirical formula found for the compound will be:

$\boxed{\boxed{C_3H_7}}\end{array}}\qquad\checkmark$

I hope this helps. =)

8 0

3 years ago

Given the following at 25C calculate delta Hf for HCN (g) at 25C. 2NH3 (g) +3O2 (g) + 2CH4 (g) ---> 2HCN (g) + 6H2O (g) delta

AysviL [449]

Answer: The $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$2NH_3(g)+3O_2(g)+2CH_4(g)\rightarrow 2HCN(g)+6H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(HCN)})+(6\times \Delta H_f_{(H_2O)})]-[(2\times \Delta H_f_{(NH_3)})+(3\times \Delta H_f_{(O_2)})+(2\times \Delta H_f_{(CH_4)})]$

We are given:

$\Delta H_f_{(H_2O)}=-241.8kJ/mol\\\Delta H_f_{(NH_3)}=-80.3kJ/mol\\\Delta H_f_{(CH_4)}=-74.6kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=-870.8kJ$

Putting values in above equation, we get:

$-870.8=[(2\times \Delta H_f_{(HCN)})+(6\times (-241.8))]-[(2\times (-80.3))+(3\times (0))+(2\times (-74.6))]\\\\\Delta H_f_{(HCN)}=135.1kJ$

Hence, the $\Delta H_f$ for HCN (g) in the reaction is 135.1 kJ/mol.

8 0

3 years ago

The electron configuration of Chromium is?

Zinaida [17]

Answer:

[Ar] 3d⁵ 4s¹ ..................

7 0

4 years ago