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Vlad1618 [11]
3 years ago
6

How many ions are there in 0.187 mol of Na+ ions

Chemistry
2 answers:
kondaur [170]3 years ago
4 0
Best Answer
1 mole of a substance contains 6.022x10^23 "units" of that substance.

So 0.187 mol of Na+ is 1.13x10^23 ions (6.022x10^23 x 0.187).
Greeley [361]3 years ago
3 0

Answer : The number of ions present in 0.187 mole of sodium ions are, 1.163\times 10^{23}

Solution : Given,

Moles of sodium ions = 0.187 moles

As we know that

1 mole of substance contains 6.22\times 10^{23} number of ions

So, 0.187 moles of substance contains 0.187\times (6.22\times 10^{23})=1.163\times 10^{23} number of ions

Therefore, the number of ions present in 0.187 mole of sodium ions are, 1.163\times 10^{23}


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when propene gas (C3H6) is added to bromine liquid, another liquid is produced which has the formula C3H6Br2. what is the chemic
klio [65]

Answer:

C3H6 + Br2 → C3H6Br2

Explanation:

The reaction in which C3H6Br2 (1,2-Dibromopropane) is created is:

  • C3H6 + Br2 → C3H6Br2

We can see that the only difference between the product (C3H6Br2) and the known reactant (C3H6) of the reaction is two bromine atoms (Br2). Br2 is diatomic bromine - a molecule we get after combining two bromine atoms. This compound is a red-brown liquid at room temperature, which means that that is the liquid described in your question.

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How could we separate the baking soda from the after after it had already been mixed in? helppp!!!
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How we can preserve the wetlands areas of nepal?<br>​
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3 years ago
Hydrocarbons, compounds containing only carbon and hydrogen, are important in fuels. The heat of combustion of cyclobutane, C4H8
Gennadij [26K]

Answer:

1. C4H8 + 6O2  -----> 4CO2 + 4H20

2. 3836.77 kcal

Explanation:

1. Balanced equation for the complete combustion of cyclobutane:

C4H8 + 6O2  -----> 4CO2 + 4H20

2. Heat of combustion of cyclobutane = 650.3 kcal/mol

    Molecular weight of cyclobutane, C4H8 = 56.1 g/mol

   Mole of C4H8 : mass of cyclobutane/Molecular weight of cyclobutane

  Mole of C4H8 = 331/56.1 = 5.9 mol

Energy released during combustion =  5.9 mol × 650.3 kcal/mol = 3836.77kcal

Therefore the energythat  is released during the complete combustion of 331 grams of cyclobutane is 3836.77kcal

8 0
3 years ago
How many moles of gas X are present if the gas has a volume of 2dm³ at room temperature and pressure? Give your answer to 2 deci
bezimeni [28]

Answer:

Approximately 0.08\; \rm mol, assuming that this gas is an ideal gas.

Explanation:

Look up the standard room temperature and pressure:25\; \rm ^{\circ}C and P = 101.325 \; \rm kPa.

The question states that the volume of this gas is V = 2\; \rm dm^{3}.

Convert the unit of all three measures to standard units:

\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}.

\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}.

\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}.

Look up the ideal gas constant in the corresponding units: R \approx 8.31\; \rm m^{3}\cdot Pa \cdot mol^{-1} \cdot K^{-1}.

Let n denote the number of moles of this gas in that V = 2\; \rm dm^{3}. By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

P \cdot V = n \cdot R \cdot T.

Rearrange this equation and solve for n:

\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}.

In other words, there is approximately 2\; \rm mol of this gas in that V = 2\; \rm dm^{3}.

6 0
3 years ago
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