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3 years ago

Mass, size, and color are examples of what?

Chemistry

1 answer:

Igoryamba3 years ago

3 0

Physical properties of matter

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Una disolución acuosa contiene el aminoácido glicina (NH2CH2COOH). Suponiendo que

yanalaym [24]

What do you mean I don’t understand this I could help but don’t understand

4 0

3 years ago

Consider the chemical equation for the production of water: 2 H2+O2→2 H2O. If 100 grams of oxygen gas are used, what would the p

Cerrena [4.2K]

Answer:

The percentage yield of water is 66.67%.

Explanation:

$2H_2+O_2\rightarrow 2H_2O$

Mass of oxygen gas = 100 g

Moles of oxygen gas = $\frac{100 g}{32 g/mol}=3.125 mol$

According to reaction, 1 mole of oxygen gives 2 moles of water, then 3.125 moles of oxygen will give:

$\frac{2}{1}\times 3.125 mol=6.25 mol$

Mass of 6.25 moles of water :

6.25 mol × 18 g/mol = 112.5 g

Theoretical yield of water = 112.5 g

Experimental yield of water = 75 g

Percentage yield :

$=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{75 g}{112.5 g}\times 100=66.67\%$

The percentage yield of water is 66.67%.

4 0

3 years ago

What is the maximum number of electrons that can go into the D orbitals

solong [7]

The maximum number is 10 electrons

3 0

3 years ago

Does lithium or potassium have a strong nuclear pull

saveliy_v [14]

Answer:

i believe it is lithium cause its is used for batteries and potassium is in like foods right? lmk if im wrong

and potassium helps to protect from radiation

Explanation:

7 0

2 years ago

A sample of oxygen gas in one container has a volume of 20.0mililiter at 297 K and 101.3 kPa. The entire sample is transferred t

VashaNatasha [74]

Answer: $V_2=\frac{101.3kPa\times 20.0ml\times 283K}{297K\times 94.6kPa}$

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=\frac{P_1V_1T_2}{T_1P_2}$

where,

$P_1$ = initial pressure of gas = 101.3 kPa

$P_2$ = final pressure of gas = 94.6 kPa

$V_1$ = initial volume of gas = 20.0 ml

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $297K$

$T_2$ = final temperature of gas = $283K$

Now put all the given values in the above equation, we get the final volume of gas.

$V_2=\frac{101.3kPa\times 20.0ml\times 283K}{297K\times 94.6kPa}$

$V_2=20.4ml$

Thus the correct numerical setup for calculating the new volume is $\frac{101.3kPa\times 20.0ml\times 283K}{297K\times 94.6kPa}$

3 0

3 years ago