I am used for making sari and I obtained from animals who am I

Pentaborane-9, B5H9, is a colorless, highly reactive liquid that will burst into flame when exposed to oxygen. The reaction is 2

mina [271]

<u>Answer:</u> The amount of energy released per gram of $B_5H_9$ is -71.92 kJ

<u>Explanation:</u>

For the given chemical reaction:

$2B_5H_9(l)+12O_2(g)\rightarrow 5B_2O_3(s)+9H_2O(l)$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

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

$\Delta H^o_{rxn}=[(5\times \Delta H^o_f_{(B_2O_3(s))})+(9\times \Delta H^o_f_{(H_2O(l))})]-[(2\times \Delta H^o_f_{(B_5H_9(l))})+(12\times \Delta H^o_f_{(O_2(g))})]$

Taking the standard enthalpy of formation:

$\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol\\\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(5\times (1271.94))+(9\times (-285.83))]-[(2\times (73.2))+(12\times (0))]\\\\\Delta H^o_{rxn}=-9078.57kJ$

We know that:

Molar mass of pentaborane -9 = 63.12 g/mol

By Stoichiometry of the reaction:

If 2 moles of $B_5H_9$ produces -9078.57 kJ of energy.

Or,

If $(2\times 63.12)g$ of $B_5H_9$ produces -9078.57 kJ of energy

Then, 1 gram of $B_5H_9$ will produce = $\frac{-9078.57kJ}{(2\times 63.12)}\times 1g=-71.92kJ$ of energy.

Hence, the amount of energy released per gram of $B_5H_9$ is -71.92 kJ

8 0

3 years ago

Monel metal is a corrosion-resistant copper-nickel alloy used in the electronics industry. A particular alloy with a density of

klemol [59]

<span>To find the volume of the plate without accounting for the hole firstly
V = (15.0 cm)(12.5 cm)(0.250 cm) = 46.875 cm^3
and the volume of the hole is
(pi)(1.25 cm)^2(0.250 cm) = 1.2272 cm^3
we will subtract the volume of the hole from the rest 45.648 cm^3
the multiply this by the density of the alloy to find the mass
(8.80 g/cm^3)(45.648 cm^3) = 401.701 g.
0.044% of this is Si, so (0.00044)(401.701 g) = 0.17675 g is silicon.
by the number of atoms and using average atomic mass of silicon and Avogadro's number to find the number of silicon atoms:
(0.17675 g)(1 mol/28.0855 g)(6.022E23 atoms/1 mol) =3.794E21atoms of Si
3.10% of these are Si-30:(0.0310)(3.794E18 atoms)=1.176E20 atoms of Si-30 and with two significant figures, 1.2E20 atoms.
hope this helps
</span>

4 0

3 years ago

= 25 X 5 = (use the correct number of sig figs)

Anton [14]

Answer:

1.25 *10^2

Explanation:

25*5 = 125

= 1.25 *10^2

5 0

3 years ago

Need help !!!!! ASAP

Ksivusya [100]

<h2>Hello!</h2>

The answer is:

We have that there were produced 0.120 moles of $CO_{2}$

$n=0.120mol$

We are asked to calculate the number of moles of the given gas, also, we are given the volume, the temperature and the pressure of the gas, we can calculate the approximate volume using The Ideal Gas Law.

The Ideal Gas Law is based on Boyle's Law, Gay-Lussac's Law, Charles's Law, and Avogadro's Law, and it's described by the following equation:

$PV=nRT$

Where,

P is the pressure of the gas.

V is the volume of the gas.

n is the number of moles of the gas.

T is the absolute temperature of the gas (Kelvin).

R is the ideal gas constant (to work with pressure in mmHg), which is equal to:

$R=62.363\frac{mmHg.L}{mol.K}$

We must remember that the The Ideal Gas Law equation works with absolute temperatures (K), so, if we are given relative temperatures such as Celsius degrees or Fahrenheit degrees, we need to convert it to Kelvin before we proceed to work with the equation.

We can convert from Celsius degrees to Kelvin using the following formula:

$Temperature(K)=Temperature(C\°) + 273K$

So, we are given the following information:

$Pressure=760mmHg\\Volume=2.965L\\Temperature=25.5C\°=25.5+273K=298.5K$

Now, isolating the number of moles, and substituting the given information, we have:

$PV=nRT$

$n=\frac{PV}{RT}$

$n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}$

$n=\frac{760mmHg*2.965L}{62.363\frac{mmHg.L}{mol.K}*298.5K}\\\\n=\frac{2242mmHg.L}{18615.355\frac{mmHg.L}{mol.}}\\\\n=0.120mole$

Hence, we have that there were produced 0.120 moles of $CO_{2}$

$n=0.120mol$

Have a nice day!

7 0

3 years ago

6CO2 + 6 H2O —> C6H12O6 + 6O2

PilotLPTM [1.2K]

Answer:

The answer to your question is 0.5 moles

Explanation:

Data

moles of Glucose = ?

moles of carbon dioxide = 3

Balanced chemical reaction

6CO₂ + 6H₂O ⇒ C₆H₁₂O₆ + 6O₂

Process

To solve this problem, use proportions, and cross multiplication.

Use the coefficients of the balanced equation.

6 moles of CO₂ ----------------- 1 mol of C₆H₁₂O₆

3 moles of CO₂ ---------------- x

x = (3 x 1) / 6

-Simplification

x = 3/6

-Result

x = 0.5 moles of Glucose

3 0

3 years ago

I am used for making sari and I obtained from animals who am I​

I am used for making sari and I obtained from animals who am I