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2 years ago

Can anyone explain this to me?

Chemistry

1 answer:

Kipish [7]2 years ago

5 0

The answer is D, the definition of standard enthalpy of formation is as follows: when one mole of a substance in the standard state (1 atm of pressure and 298.15 K) is formed from its pure elements. The last two words being especially important to this question. We would rule out A, B and C all because the product is formed from a mixture of an element and a compound and not just of elements. In other questions the reactants may all be in elemental form so you must look at the products being one mole or the reactants not being in their standard state.

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Doctors use the radioactive isotope chromium-51 to label red blood cells in the human body. Chromium-51 gives off relatively ___

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Alpha particles are relatively heave and can be stopped by a sheet of paper.

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3 years ago

Read 2 more answers

The fuel used to power the booster rockets on

Effectus [21]

Answer:

746 moles of H2O are been produced from 373 moles of Al.

Explanation:

For every 3 moles of aluminum, you get 6 moles of H2O (double). Therefore, every 373 moles of Al, you will get double as well, that is 746 m.

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3 years ago

Calculate the numbers of mL in 0.603 L

Artist 52 [7]

Answer:

603 mL

Explanation:

A milliliter is a unit of volume equal to 1/1000th of a liter. It is the same as a cubic centimeter.

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3 years ago

The volume of an automobile air bag was 66.8 L when inflated at 25 °C with 77.8 g of nitrogen gas. What was the pressure in the

Ray Of Light [21]

Answer:

$\large \boxed{\text{103 kPa}}$

Explanation:

We can use the Ideal Gas Law — pV = nRT

Data:

V = 66.8 L

m = 77.8 g

T = 25 °C

Calculations:

(a) Moles of N₂

$\text{Moles of N}_{2} = \text{77.8 g N}_{2} \times \dfrac{\text{1 mol N}_{2}}{\text{28.01 g N}_{2}} = \text{2.778 mol N}_{2}$

(b) Convert the temperature to kelvins

T = (25 + 273.15) K = 298.15 K

(c) Calculate the pressure

$\begin{array}{rcl}pV & =& nRT\\p \times \text{66.8 L} & = & \text{2.778 mol} \times \text{8.314 kPa$\cdot$ L$\cdot$K$^{-1}$mol$^{-1}\times$ 298.15 K}\\66.8p & = & \text{6886 kPa}\\p & = & \textbf{103 kPa}\end{array}\\\text{The pressure in the bag is $\large \boxed{\textbf{103 kPa}}$}$

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2 years ago

You are performing a titration of a triprotic acid, when you spill water on your lab notebook. you can read that: pka 1 = 1.40,

eimsori [14]

According to the PH formula:
PH= Pka +㏒ [strong base/weak acid]
when we have PH at the first equivalence =3.35 and the Pka1 = 1.4
So, by substitution, we can get the value of ㏒[strong base / weak acid]
3.35 = 1.4 + ㏒[strong base/ weak acid]
∴㏒[strong base/weak acid] = 3.35-1.4 = 1.95
to get the Pka2 we will substitute with the value of ㏒[strong base/ weak acid] and the value of PH of the second equivalence point
∴Pk2 = PH2 - ㏒[strong base/ weak acid]
= 7.55 - 1.95 = 5.6

5 0

3 years ago