Please help me I really don’t understand

What is the formula weight (amu) of the molecule H2O? Use atomic masses of H and O as 1.008 amu and 16.00 amu respectively. Repo

vaieri [72.5K]

Answer:

Formula weight of H₂O molecule is 18.02 amu.

Explanation:

Given data:

Formula weight of H₂O = ?

Atomic mass of H = 1.008 amu

Atomic mass of O = 16.00 amu

Solution:

Formula weight:

"It is the sum of all the atomic weight of atoms present in given formula"

Formula weight of H₂O = 2×1.008 amu + 1×16.00 amu

Formula weight of H₂O = 18.02 amu

Thus, formula weight of H₂O molecule is 18.02 amu.

6 0

3 years ago

What volume of O2 collected at 22.0 and 728 mmHg would be produce by the decomposition of 8.15 g KClO3?

adell [148]

Answer:

There is 2.52 L of O2 collected

Explanation:

Step 1: Data given:

Temperature = 22.0 °C

Pressure = 728 mmHg = 728 /760 = 0.958 atm

Mass of KClO3 = 8.15 grams

Molar mass of KClO3 = 122.55 g/mol

Step 2: The balanced equation

2KClO3(s) → 2KCl(s) + 3O2(g)

Step 3: Calculate moles of KClO3

Moles KClO3 = mass KClO3 / molar mass KClO3

Moles KClO3= 8.15 grams / 122.55 g/mol

Moles KClO3 = 0.0665 moles

Step 4: Calculate moles of O2

For 2 moles of KClO3 we'll have 2 moles of KCl and 3 moles of O2 produced

For 0.0665 moles of KClO3 we have 3/2 * 0.0665 = 0.09975 moles

Step 5: Calculate vlume of O2

p*V = n*R*T

V = (n*R*T)/p

⇒ with n = the number of moles O2 = 0.09975 moles

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = 22.0 °C = 273 +22 = 295 Kelvin

⇒ with p = 0.958 atm

V = (0.09975 * 0.08206 * 295) / 0.958

V = 2.52 L

There is 2.52 L of O2 collected

7 0

3 years ago

The radioisotope phosphorus-32 is used in tracers for measuring phosphorus uptake by plants. The half-life of phosphorus-32 is 1

Harman [31]

Answer:

54 days

Explanation:

We have to use the formula;

0.693/t1/2 =2.303/t log Ao/A

Where;

t1/2= half-life of phosphorus-32= 14.3 days

t= time taken for the activity to fall to 7.34% of its original value

Ao=initial activity of phosphorus-32

A= activity of phosphorus-32 after a time t

Note that;

A=0.0734Ao (the activity of the sample decreased to 7.34% of the activity of the original sample)

Substituting values;

0.693/14.3 = 2.303/t log Ao/0.0734Ao

0.693/14.3 = 2.303/t log 1/0.0734

0.693/14.3 = 2.6/t

0.048=2.6/t

t= 2.6/0.048

t= 54 days

3 0

3 years ago

Calculate the volume occupied by 272g

cupoosta [38]

The answer for the following problem is mentioned below.

Therefore volume occupied by methane gas is 184.78 × 10^-3 liters

Explanation:

Given:

mass of methane( $CH_{4}$ ) = 272 grams

pressure (P) = 250 k Pa =250×10^3 Pa

temperature(t) = 54°C =54 + 273 = 327 K

Also given:

R = 8.31JK-1 mol-1 ,

Molar mass of methane( $CH_{4}$ ) = 16.0 grams

We know;

According to the ideal gas equation,

P × V = n × R × T

here,

n = m÷M

n =272 ÷ 16

n = 17 moles

Therefore,

250×10^3 × V = 17 × 8.31 × 327

V = ( 17 × 8.31 × 327 ) ÷ ( 250×10^3 )

V = 184.78 × 10^-3 liters

Therefore volume occupied by methane gas is 184.78 × 10^-3 liters

6 0

3 years ago

A 45-g aluminium spoon(specific heat 0.80 / J/gdegree Celsius) at 24 degree celsius is placed in 180 ml(180 grams) of coffee at

Vlad [161]

Explanation:

a) The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, $heat_{absorbed}=heat_{released}$

To calculate the amount of heat released or absorbed, we use the equation:

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

Also,

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ..........(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 45 g

$m_2$ = mass of coffee = 180 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminium = $24^oC$

$T_2$ = temperature of coffee = $85^oC$

$c_1$ = specific heat of aluminium = $0.80J/g^oC$

$c_2$ = specific heat of coffee= $4.186 J/g^oC$

Putting all the values in equation 1, we get:

$45 g\times 0.80J/g^oC\times (T_{final}-24^oC)=-[180 g\times 4.186J/g^oC\times (T_{final}-83^oC)]$

$T_{final}=80.30^oC$

80.30 °C is the final temperature.

b) Energy flows from higher temperature to lower temperature.Whenever two bodies with different energies and temperature come in contact. And the resulting temperature of both bodies will less then the body with high temperature and will be more then the body with lower temperature.

So, is our final temperature of both aluminium and coffee that is 80°C less than initial temperature of coffee and more than the initial temperature of the aluminum.

8 0

3 years ago