To determine the pressure in units of kPa, we need to use a conversion factor to convert the units from mmHg to kPa. A conversion factor is a value that would relate two different units and is multiplied or divide to the original measurement depending on what is units is asked. From literature, 1 atm is equal to 760 mmHg and it is also equal to 101.325 kPa. We use these factors to convert the given value. We do as follows:

2150 mmHg ( 1 atm / 760 mmHg ) ( 101.325 kPa / 1 atm ) = 286.643 kPa

Therefore, the closest value from the choices is the second one which has the value of 287, this would be answer.

7 0

3 years ago

Beryllium oxide, Beo, is an electrical insulator. How

egoroff_w [7]

Answer:

There are 10.0 moles of beryllium oxide in a 250 grams sample of the compound.

Explanation:

We can calculate the number of moles (η) of BeO as follows:

$\eta = \frac{m}{M}$

Where:

m: is the mass = 250 g

M: is the molar mass = 25.0116 g/mol

Hence, the number of moles is:

$\eta = \frac{250 g}{25.0116 g/mol} = 10.0 moles$

Therefore, there are 10.0 moles of beryllium oxide in a 250 grams sample of the compound.

I hope it helps you!

3 0

3 years ago

This week in lab you will use the spectrovis plus spectrophotometer to measure the light that is absorbed by dopachrome, a dark

NARA [144]

A solution that appears dark orange in color is not absorbing all orange wavelengths of light.

6 0

3 years ago

The formation of ammonia is represented by the equation N2(g) + 3H2(g) ⇌ 2NH3(g). Determine the enthalpy of formation of ammonia

Savatey [412]

Answer:

$\Delta _fH_{NH_3}=-66\frac{kJ}{mol}$

Explanation:

Hello!

In this case, since the study of the bond energy allows us to compute the enthalpies of some reactions, for this combination reaction by which ammonia is yielded, we understand the enthalpy of reaction equals the enthalpy of formation of ammonia, and, in terms of the bonds energy we can write:

$\Delta _fH_{NH_3}=Delta _rH=\Sigma \Delta H(bonds \ broken)-\Sigma \Delta H(bonds \ formed)$

Whereas the bonds enthalpy of those bonds that get broken cover the N≡N and the three H-H bonds at the reactants side and the enthalpy of those bonds that are formed cover the six N-H bonds at the products; which means we obtain:

$\Delta _fH_{NH_3}=942\frac{kJ}{mol} +3*436\frac{kJ}{mol}-6*386\frac{kJ}{mol}\\\\\Delta _fH_{NH_3}=-66\frac{kJ}{mol}$

Which differs from the theoretical value that is -46 kJ/mol.

Best regards!

7 0

3 years ago