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Alexeev081 [22]

3 years ago

11

What is the density of 1.00 mol of carbon monoxide gas at STP?

1 answer:

Ksju [112]3 years ago

6 0

Answer:

Density = mass / volume.

If this is an ideal gas then 1mol will take up 22.4L of volume (fact about ideal gases you should remember)

Since you have 1mol then you know the volume of the gas that you have (22.4L)

Now, just convert 1mol of CO to grams.The ptable tells you that the mass of 1 mol of C is 12g and the mass of 1 mol of O is 16g. So the mass of 1 mol of CO is......... :):)

Now you have the mass and the volume, so just divide :) enjoy

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Which set of quantum numbers is not permitted by the rules of quantum mechanics?

sineoko [7]

Only these values can exist... N=? L=0 to N-1 M=+ to - L S=+ or - 1/2

3 0

3 years ago

Read 2 more answers

If mass =180kg and volume =90m3, what is the density

DaniilM [7]

To find the density you must divide the mass by the density.
180kg ÷ 90m³ = 2kg/m<span>³
The density is </span>2kg/m³

4 0

3 years ago

Calculate the mass of water formed when 4g of methane is burnt in excess oxygen

wariber [46]

Answer:

Mass of water formed: 9g

Explanation:

Hope it helps you!

6 0

3 years ago

A 50.0 g sample of liquid water at 25.0 degree C is mixed with 29.0 g of water at 45 degree C. The final temperature of the wate

kotegsom [21]

<u>Answer:</u> The final temperature of water is 32.3°C

<u>Explanation:</u>

When two solutions are mixed, the amount of heat released by solution 1 (liquid water) will be equal to the amount of heat absorbed by solution 2 (liquid water)

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of solution 1 (liquid water) = 50.0 g

$m_2$ = mass of solution 2 (liquid water) = 29.0 g

$T_{final}$ = final temperature = ?

$T_1$ = initial temperature of solution 1 = 25°C = [273 + 25] = 298 K

$T_2$ = initial temperature of solution 2 = 45°C = [273 + 45] = 318 K

c = specific heat of water= 4.18 J/g.K

Putting values in equation 1, we get:

$50.0\times 4.18\times (T_{final}-298)=-[29.0\times 4.18\times (T_{final}-318)]\\\\T_{final}=305.3K$

Converting this into degree Celsius, we use the conversion factor:

$T(K)=T(^oC)+273$

$305.3=T(^oC)+273\\T(^oC)=(305.3-273)=32.3^oC$

Hence, the final temperature of water is 32.3°C

7 0

3 years ago

The addition of hydrochloric acid to a silver nitrate solution precipitates silver chloride according to the reaction:

Nimfa-mama [501]

Answer:

Enthalpy change for the reaction is -67716 J/mol.

Explanation:

Number of moles of $AgNO_{3}$ in 50.0 mL of 0.100 M of $AgNO_{3}$

= Number of moles of HCl in 50.0 mL of 0.100 M of HCl

= $\frac{0.100}{1000}\times 50.0$ moles

= 0.00500 moles

According to balanced equation, 1 mol of $AgNO_{3}$ reacts with 1 mol of HCl to form 1 mol of AgCl.

So, 0.00500 moles of $AgNO_{3}$ react with 0.00500 moles of HCl to form 0.00500 moles of AgCl

Total volume of solution = (50.0+50.0) mL = 100.0 mL

So, mass of solution = ( $100.0\times 1.00$ ) g = 100 g

Enthalpy change for the reaction = -(heat released during reaction)/(number of moles of AgCl formed)

= $\frac{-m_{solution}\times C_{solution}\times \Delta T_{solution}}{0.00500mol}$

= $\frac{-100g\times 4.18\frac{J}{g.^{0}\textrm{C}}\times [24.21-23.40]^{0}\textrm{C}}{0.00500mol}$

= -67716 J/mol

[m = mass, c = specific heat capacity, $\Delta T$ = change in temperature and negative sign is included as it is an exothermic reaction]

4 0

3 years ago