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olya-2409 [2.1K]
3 years ago
15

Calculate the number of moles in 1.8gram of h2o​

Chemistry
1 answer:
Rzqust [24]3 years ago
7 0

Answer:

~.1058 Moles

Explanation:

The formula for this question is the following.

1 mole of a compound/molar mass of the compound.

First we need to find the molar mass of H2O, which is the atomic mass on the periodic table. Hydrogen is 1.01, Oxygen is 16.00. Add those together to get the molar mass of the compound and you'll get an equation that looks like this.

1 mole of H2O/ 17.01 g/mol H2O

We now know that in 1 mole of H20 there is 17.01 g.

Take 1.8g and divide it by 17.01, you get your answer.

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<span>There are 2.74603 * 10^25 hydrogen atoms. Ammonium sulfide is represented by (NH4)2S. This means that there are 8 hydrogen atoms total. There are also 8 mol of H in ammonium sulfide. We also need to use avogadro’s number of 6.022 * 10^23. Hydrogen Atoms = 5.7mol * (8mol / 1mol) * 6.022 * 10^23 per mol Hydrogen atoms = 2.74603 * 10^25 hydrogen atoms.</span>
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Why does calcium conduct electricity?
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calcium is a metal and metals are good conductors of electricity as they contain mobile electrons.

Explanation:

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Which of the following does NOT play a role in South Florida’s climate?
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i think a im not sure

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3 years ago
The chemical equation of rusting of iron is given
artcher [175]

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{ \tt{(a). { \bf{reactants : { \tt{iron \:  and \: oxygen}}}}}} \\ { \bf{products :  { \tt{iron(iii)oxide}}}} \\  \\ { \tt{(b).}} \: { \bf{1.204 \times  {10}^{24}  \: atoms}} \\ { \tt{(c).}} \: { \tt{4Fe +3 O _{2}  → 2Fe _{2}O _{3}}} \\  \\ { \underline{ \blue{ \tt{becker \: jnr}}}}

6 0
2 years ago
When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C
natka813 [3]

Answer:

\Delta_{r}U of the reaction is -6313 kJ/mol

\Delta_{r}H of the reaction is -6312 kJ/mol

Explanation:

Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C

q_{cal}= C \times \Delta T

=5.861 \times 3.596 = 21.076\,kJ

q_{rxn}= -q_{cal}= -21.076\,kJ

\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol

Therefore, \Delta_{r}U of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter  is as follows.

C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)

Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5

\Delta_{r}H=\Delta E+ \Delta n. RT

=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol

Therefore, \Delta_{r}H of the reaction is -6312 kJ/mol.

3 0
3 years ago
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