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2 years ago

If 32.0 g of MgSO4⋅7H2O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?

1 answer:

6 0

Find moles of MgSO4.7H2O

molar mass = 246
so moles = 32 / 246 = 0.13 moles.

When heated, all 7 H2O from 1 molecule will be gone.
total moles of H2O present = 7 x 0.13 = 0.91
mass of those H2O = 0.91 x 18 = 16.38g

so mass of anyhydrous MgSO4 remain = 32 - 16.38 = 15.62 g

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The concentration of C29H60 in summer rainwater is 34 ppb. Find the molarity of this compound in nanomoles per liter (nM).

Naya [18.7K]

Answer: The concentration of C29H60 in nM per liter is 83,33 nM/liter

Explanation: Let's start from the ppb definition: ppb means parts per billion. In terms of concentracion measuring this means micrograms of solute per liter of solution.

The algebraic expression would be:

ppb [=] micrograms of compound/liter of solution

We can assume that the solvent is water. The solute is dissolved in water and both create the C29H60 solution.

For the exercise we have 34 ppb of C29H60, that means 34 micrograms of C29H60 in one liter of solution. So, since now, we have to convert the units from the initial data to the required answer.

The respective procedure is in a attached file.

5 0

1 year ago

A sample of 8.5 g nh3 on oxidation produces 4.5 g of no. calculate the percent yield. reaction: 4 nh3 5 o2 → 4 no 6 h2o

AnnZ [28]

30.% Determine the molar mass of NH3 and NO Atomic weight nitrogen = 14.0067 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Molar mass NH3 = 14.0067 + 3 * 1.00794 = 17.03052 g/mol Molar mass NO = 14.0067 + 15.999 = 30.0057 g/mol Moles NH3 = 8.5 / 17.03052 = 0.499103962 mol Moles NO = 4.5 / 30.0057 = 0.149971505 mol Looking at the balanced equation, for every mole of NH3 consumed, you should get one mole of NO. So if we had 100% yield, we should have 0.499103962 moles of NO. But we don't. The percent yield is a simple matter of division by what we did get by what we should get. So 0.149971505 / 0.499103962 = 0.300481497 = 30.0481497% Rounding to 2 significant figures gives 30.% yield.

5 0

2 years ago

An atom of lead has a radius of and the average orbital speed of the electrons in it is about . Calculate the least possible unc

victus00 [196]

The question is incomplete. Here is the complete question.

An atom of lead has a radius of 154 pm and the average orbitalspeed of the electron in it is about 1.8x $10^{8}$ m/s. Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of lead. Write your answer as a percentage of the average speed, and round it to significant 2 digits.

Answer: v% = 0.21 m/s

Explanation: To calculate the uncertainty, use Heisenberg's Uncertainty Principle, which states that: ΔpΔx≥ $\frac{h}{4\pi }$

where h is Planck's constant and it is equal to 6.626. $10^{-34}$ m²kg/s.

Since p (momentum) is p = m.v:

mΔv.Δx ≥ $\frac{h}{4\pi }$

Δv = $\frac{h}{4\pi.x.m }$

Given that: r = x = 1.54. $10^{-10}$ m and mass of an electron is m=9.1. $10^{-31}$ kg

Δv = $\frac{6.626.10^{-34} }{4.3.14.1.54.10^{-10}.9.1.10^{-31}}$

Δv = 0.0376. $10^{7}$

As percentage of average speed:

Δv. $\frac{1}{v}$ .100% = $\frac{0.0376.10^{7} }{1.8.10^{8} }$ .10² = 0.021.10 = 0.21%

The least possible uncertainty in a speed of an electron is 0.21%.

5 0

2 years ago

a drop of water weighing 0.48 g condenses on the surface of a 55-g block of aluminum that is initially at 25C. if the heat durin

Gwar [14]

Mass of water vapor is 0.48 gms.

Weight of the aluminium block is 55 gms.

Heat of vaporization of water at 25 degree Celsius is 44.0 kJ/mol.

The amount of heat given by the condensation is:

q = Heat of vaporization × Mass of vapor / Molar mass of steam

= 44.0 kJ/mol ×0.48 g / 18 g/mol

= 1.173 kJ

Now the final temperature of the metal block is calculated by the formula:

q = m × c × ΔT

q = m × c × (T₂ - T₁)

Here, q is the amount of heat, m is the mass of the metal, c is the specif heat of the metal, T₁ is initial temperature, and T₂ is the final temperature.

Now, substituting the values we get,

1.173 kJ = 55 g × 0.903 J/g. ° C * (T₂ - 25°C)

1.173 × 10³ J = 55 g × 0.903 J/g. degree C × (T₂ - 25°C)

1.173 × 10³ = 49.665 ° C * (T₂ - 25° C)

T₂ = 49° C.

Thus, the final temperature of the metal block is 49° C.

8 0

2 years ago

Find the age t of a sample, if the total mass of carbon in the sample is mc, the activity of the sample is a, the current ratio

kkurt [141]

should be 7.57*10^3 years

4 0

2 years ago