Calculate q (in units of joules) when 1.850 g of water is heated from 22 °C to 33 °C

If 16.00 g of O₂ reacts with 80.00 g NO, how many grams of NO₂ are produced? (enter only the value, round to whole number)

Norma-Jean [14]

Answer:

46 g

Explanation:

The balanced equation of the reaction between O and NO is

2 NO + O₂ ⇔ 2 NO₂

Now, you need to find the limiting reagent. Find the moles of each reactant and divide the moles by the coefficient in the equation.

NO: (80 g)/(30.006 g/mol) = 2.666 mol

(2.666 mol)/2 = 1.333

O₂: (16 g)/(31.998 g/mol) = 0.500 mol

(0.500 mol)/1 = 0.500 mol

Since O₂ is smaller, this is the limiting reagent.

The amount of NO₂ produced will depend on the limiting reagent. You need to look at the equation to determine the ratio. For every mole of O₂ reacted, 2 moles of NO₂ are produced.

To find grams of NO₂ produced, multiply moles of O₂ by the ratio of NO₂ to O₂. Then, convert moles of NO₂ to find grams.

0.500 mol O₂ × (2 mol NO₂/1 mol O₂) = 1.000 mol NO₂

1.000 mol × 46.005 g/mol = 46.005 g

You will produce 46 g of NO₂.

6 0

3 years ago

Both isopropyl alcohol, C3H8OH, and ethylene glycol, C2H6O2, are used as antifreeze. When equal masses of each are added to wate

kiruha [24]

I believe the correct answer from the choices listed above is option B. When equal masses of each are added to water, ethylene glycol would be more effective. Ethylene glycol is the most widely used automotive cooling-systemantifreeze, although methanol, ethanol, isopropyl alcohol, and propylene glycol are also used.
Hope this answers the question. Have a nice day.

6 0

3 years ago

Read 2 more answers

Which processes represent one chemical change in one physical change

prisoha [69]

decomposition and melting

7 0

3 years ago

Read 2 more answers

Be sure to answer all parts. In the average adult male, the residual volume (RV) of the lungs, the volume of air remaining after

Alenkinab [10]

Answer:

For a: The number of moles of air present in the RV is 0.047 moles

For b: The number of molecules of gas is $2.83\times 10^{22}$

Explanation:

For a:

To calculate the number of moles, we use the equation given by ideal gas follows:

$PV=nRT$

where,

P = pressure of the air = 1.00 atm

V = Volume of the air = 1200 mL = 1.2 L (Conversion factor: 1 L = 1000 mL)

T = Temperature of the air = $37^oC=[37+273]K=310K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of air = ?

Putting values in above equation, we get:

$1.00atm\times 1.2L=n_{air}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 310K\\n_{air}=\frac{1.00\times 1.2}{0.0821\times 310}=0.047mol$

Hence, the number of moles of air present in the RV is 0.047 moles

For b:

According to mole concept:

1 mole of a compound contains $6.022\times 10^{23}$ number of molecules.

So, 0.047 moles of air will contain $(0.047\times 6.022\times 10^{23})=2.83\times 10^{22}$ number of gas molecules.

Hence, the number of molecules of gas is $2.83\times 10^{22}$

7 0

3 years ago

A

BARSIC [14]

The number of C atoms in 0.524 moles of C is 3.15 atoms.

The number of $SO_2$ molecules in 9.87 moles $SO_2$ is 59.43 molecules.

The moles of Fe in 1.40 x $10^{22}$ atoms of Fe is 0.23 x $10^{-1}$

The moles of $C_2H_6O$ in 2.30x $10^{24}$ molecules of $C_2H_6O$ is 3.81.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × $10^{23}$ of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

A. The number of C atoms in 0.524 mole of C:

6.02214076 × $10^{23}$ x 0.524 mole

3.155601758 atoms =3.155 atoms

B. The number of $SO_2$ molecules in 9.87 moles of $SO_2$ :

6.02214076 × $10^{23}$ x 9.87

59.4385293 molecules= 59.43 molecules

C. The moles of Fe in 1.40 x $10^{22}$ atoms of Fe:

1.40 x $10^{22}$ ÷ 6.02214076 × $10^{23}$

0.2324754694 x $10^{-1}$ moles.

0.23 x $10^{-1}$ moles.

D. The moles of $C_2H_6O$ in 2.30x $10^{24}$ molecules of $C_2H_6O$ :

2.30x $10^{24}$ ÷ 6.02214076 × $10^{23}$

3.819239854 moles=3.81 moles

Learn more about moles here:

brainly.com/question/8455949

#SPJ1

6 0

2 years ago