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3 years ago

Find the area of a circle inscribed in an equilateral triangle of side 18 cm. ( Take pi= 3.14). answer should be 54.78 cm²

Chemistry

1 answer:

rodikova [14]3 years ago

8 0

Answer:

The answer according to me.............

Explanation:

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What is the final temperature of the metal

Natasha_Volkova [10]

Answer:

There is no exact answer for this question tbh.

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3 years ago

When 2.00 g of methane are burned in a bomb calorimeter, the change in temperature is 3.08°C. The heat capacity of the calorimet

melisa1 [442]

Answer:

The approximate molar enthalpy of combustion of this substance is -66 kJ/mole.

Explanation:

First we have to calculate the heat gained by the calorimeter.

$q=c\times \Delta T$

where,

q = Heat gained = ?

c = Specific heat = $2.68 kJ/^oC$

ΔT = The change in temperature = 3.08°C

Now put all the given values in the above formula, we get:

$q=2.68 kJ/^oC\times 3.08^oC$

$q=8.2544 kJ$

Now we have to calculate molar enthalpy of combustion of this substance :

$\Delta H_{comb}=-\frac{q}{n}$

where,

$\Delta H_{comb}$ = enthalpy change = ?

q = heat gained = 8.2544kJ

n = number of moles methane = $\frac{\text{Mass of methane}}{\text{Molar mass of methane }}=\frac{2.00 g}{16.042 g/mol}=0.1247 mole$

$\Delta H_{comb}=-\frac{8.2544 kJ}{0.1247 mole}=-66.21 kJ/mole\approx -66 kJ/mole$

Therefore, the approximate molar enthalpy of combustion of this substance is -66 kJ/mole.

6 0

3 years ago

When air bags inflate, nitrogen gas is formed from sodium azide, along with solid sodium. what reaction category is this?

adell [148]

The answer is
alkaline silocate

7 0

3 years ago

Helppppp with this pleasee

vazorg [7]

Answer:

A. 1, 2, 5

Explanation:

Count the number of Ns in the formula.

- Hope that helped! Please let me know if you need a further explanation.

3 0

3 years ago

You need to prepare a solution with a specific concentration of Na+Na+ ions; however, someone used the end of the stock solution

babymother [125]

Explanation:

Ionic equation

NaCl(aq) --> Na+(aq) + Cl-(aq)

Na2SO4(aq) --> 2Na+(aq) + SO4^2-(aq)

In NaCl solution, 1 mole of Na+ is dissociated in 1 liter of solution while in Na2SO4, 2 moles of Na+ is dissociated in 1 liter of solution.

Molecular weight of NA2SO4 = (23*2) + 32 + (16*4)

= 142 g/mol

Molecular weight of NaCl = 23 + 35.5

= 58.5 g/mol

Masses

% Mass of NA+ in Na2SO4 = mass of Na+/total mass of Na2SO4 * 100

= 46/142 * 100

= 32.4%

% Mass of NA+ in NaCl = mass of Na+/total mass of NaCl * 100

= 23/58.5 * 100

= 39.3%

Therefore, the % mass of Na+ in NaCl and Na2SO4 are different so it cannot be used.

7 0

2 years ago

Find the area of a circle inscribed in an equilateral triangle of side 18 cm. ( Take pi= 3.14).​ answer should be 54.78 cm²

Find the area of a circle inscribed in an equilateral triangle of side 18 cm. ( Take pi= 3.14). answer should be 54.78 cm²