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kap26 [50]
3 years ago
10

The table below gives the atomic mass and relative abundance values for the three isotopes of element M.

Chemistry
1 answer:
Serhud [2]3 years ago
7 0

The average atomic mass formula is:

average atomic mass = \Sigma (percent abundane)\times (atomic mass)

Substituting the values in the formula:

average atomic mass = \frac{78.99\times 23.9850+10\times 24.9858+11.01\times 25.9826}{100}

average atomic mass = \frac{1894.57515+249.858+286.068426}{100}

average atomic mass = 24.30516 amu

Hence, 24.30516 amu is the average atomic mass of element M.

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High temperature increases the number of high energy collisions

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Convert 125m^3 to km^3
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1.25e-7

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3 years ago
How many grams are in 23.0 moles of carbon (C)?
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276 grams of carbon in 23.0 moles
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3 years ago
Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly
trapecia [35]

Answer : The mass of silver sulfadiazine produced can be, 71.35 grams.

Solution : Given,

Mass of Ag_2O = 25.0 g

Mass of C_{10}H_{10}N_4SO_2 = 50.0 g

Molar mass of Ag_2O = 231.7 g/mole

Molar mass of C_{10}H_{10}N_4SO_2 = 250.3 g/mole

Molar mass of AgC_{10}H_{9}N_4SO_2 = 357.1 g/mole

First we have to calculate the moles of Ag_2O and C_{10}H_{10}N_4SO_2.

\text{ Moles of }Ag_2O=\frac{\text{ Mass of }Ag_2O}{\text{ Molar mass of }Ag_2O}=\frac{25.0g}{231.7g/mole}=0.1079moles

\text{ Moles of }C_{10}H_{10}N_4SO_2=\frac{\text{ Mass of }C_{10}H_{10}N_4SO_2}{\text{ Molar mass of }C_{10}H_{10}N_4SO_2}=\frac{50.0g}{250.3g/mole}=0.1998moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

Ag_2O(s)+2C_{10}H_{10}N_4SO_2(s)\rightarrow 2AgC_{10}H_9N_4SO_2(s)+H_2O(l)

From the balanced reaction we conclude that

As, 2 mole of C_{10}H_{10}N_4SO_2 react with 1 mole of Ag_2O

So, 0.1998 moles of C_{10}H_{10}N_4SO_2 react with \frac{0.1998}{2}=0.0999 moles of Ag_2O

From this we conclude that, Ag_2O is an excess reagent because the given moles are greater than the required moles and C_{10}H_{10}N_4SO_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of AgC_{10}H_9N_4SO_2

From the reaction, we conclude that

As, 2 mole of C_{10}H_{10}N_4SO_2 react with 2 mole of AgC_{10}H_9N_4SO_2

So, 0.1998 mole of C_{10}H_{10}N_4SO_2 react with 0.1998 mole of AgC_{10}H_9N_4SO_2

Now we have to calculate the mass of AgC_{10}H_9N_4SO_2

\text{ Mass of }AgC_{10}H_9N_4SO_2=\text{ Moles of }AgC_{10}H_9N_4SO_2\times \text{ Molar mass of }AgC_{10}H_9N_4SO_2

\text{ Mass of }AgC_{10}H_9N_4SO_2=(0.1998moles)\times (357.1g/mole)=71.35g

Therefore, the mass of silver sulfadiazine produced can be, 71.35 grams.

8 0
3 years ago
How many grams of O are in 605 g of Na,O?
Elden [556K]

Answer:

2Na2O2+2H2O⟶O2+4NaOH

2×78g                           32g

156g of Na2O2 produces 32g of O2,

12g of Na2O2 produces =15632×12=10.66g.

Density of O2 at NTP=1.428g/mL.

DensityMass=Vol.

1.42810.66=7.46mL

Vol. of O2 at NTP is 7.46mL.

Explanation:

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6 0
2 years ago
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