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lukranit [14]
3 years ago
14

Answer the question please

Chemistry
1 answer:
emmainna [20.7K]3 years ago
3 0
It’s 190 calories without milk
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Calculate the volume of a balloon that can hold 113.4 g of nitrogen dioxide, NO2 gas at STP-
Karolina [17]

Answer:

55.18 L

Explanation:

First we convert 113.4 g of NO₂ into moles, using its molar mass:

  • 113.4 g ÷ 46 g/mol = 2.465 mol

Then we<u> use the PV=nRT formula</u>, where:

  • P = 1atm & T = 273K (This means STP)
  • n = 2.465 mol
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • V = ?

Input the data:

  • 1 atm * V = 2.465 mol * 0.082atm·L·mol⁻¹·K⁻¹ * 273 K

And <u>solve for V</u>:

  • V = 55.18 L
6 0
2 years ago
Escriba 4 ideas sobre que sucede cuando la radiación solar llega a la tierra​
poizon [28]

Answer:

La radiación UV-B es parcialmente absorbida por el ozono y llega a la superficie de la tierra, produciendo daño en la piel. Ello se ve agravado por el agujero de ozono que se produce en los polos del planeta.

Explanation:

3 0
2 years ago
7.00 of Compound x with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25c.
beks73 [17]

Answer:

\Delta H_{f,C_3H_4}=276.8kJ/mol

Explanation:

Hello!

In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

\Delta H_{rxn} =- m_wC_w\Delta T

We plug in the mass of water, temperature change and specific heat to obtain:

\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ

Now, this enthalpy of reaction corresponds to the combustion of propyne:

C_3H_4+4O_2\rightarrow 3CO_2+2H_2O

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol

Now, we solve for the enthalpy of formation of C3H4 as shown below:

\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol

Best regards!

7 0
3 years ago
A 7.36 g sample of copper is contaminated with a additional 0.51 g sample of zinc. suppose an atomic mass measurement was perfor
OleMash [197]
The average molecular weight of the mixture can be calculated using this formula:
MWav = x1MW1 + x2MW2

Where x is the mass fraction of the components of the mixture, in this case, copper (63.546 g/mol) and zinc (<span>65.38 g/mol).
</span> 
x1 = 7.36 / (7.36+0.51)=0.935
x2 = 0.51 / (7.36+0.51)=0.065

So,
MWav = 0.935(63.546) + 0.065(65.38) = 63.665 g/mol
3 0
3 years ago
Why do metals lose shine but gold<br> does not?
morpeh [17]
Because <span>Gold is a very inert metal, and won't really react with oxygen in the air.

The only reason other metals tarnish, such as copper, silver, steel/iron, is

because they're freely reacting with air, to produce a metal oxide surface layer. </span>
3 0
3 years ago
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