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3 years ago

What is the volume, in mL, of 0.50 moles of ammonia (NH3) at STP?

1 answer:

5 0

For the purpose we will here use the ideal gas law:

p×V=n×R×T

V= ?
n = 0.5 moleT= 273.15 K (at STP)
p= 101.325 kPa (at STP)
R is universal gas constant, and its value is 8.314 J/mol×K

Now when we have all necessary date we can calculate the number of moles:
V=nxRxT/p
V=0.5x8.314x273.15/101.325= 11.2 L = 11200 mL

Answer: D.

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A water treatment tablet contains 20.0 mg of tetraglycine hydroperiodide, 40.0% of which is available as soluble iodine. If two

Galina-37 [17]

Answer:

$16\ \text{ppm}$

Explanation:

Mass of one tablet = 20 mg

Mass of two tablets = $2\times 20=40\ \text{mg}$

Percent that is soluble in water = 40%

Mass of tablet that is soluble in water = $0.4\times 40=16\ \text{mg}$

So, mass of solute is $16\ \text{mg}$

Density of water = 1 kg/L

Volume of water = 1 L

So, mass of 1 L of water is $1\times 1=1\ \text{kg}=1000\ \text{g}$

PPM is given by

$\dfrac{\text{Mass of solute}}{\text{Mass of solvent}}\times 10^6=\dfrac{16\times 10^{-3}}{1000}\times 10^6\\ =16\ \text{ppm}$

Hence, the concentration of iodine in the treated water $16\ \text{ppm}$ .

8 0

3 years ago

There is water on the pan of the scale as you measure the mass of an object. If you were to ignore the water, what would be the

mihalych1998 [28]

Remember that density refers to the "mass per unit volume" of an object.

So, if an object had a mass of 100 grams and a volume of 100 milliliters, the density would be 100 grams / 100 ml.

In the question, water on the surface of the scale would add weight, so the mass of the object that you're weighing would appear to be heavier than it really is. If that happens, you'll incorrectly assume that the density is GREATER than it really is

As an example, suppose that there was 5 ml of water on the surface of the scale. Water has a density of 1 gram per milliliter (1 g/ml) so the water would add 5 grams to the object's weight. If we use the example above, the mass of the object would seem to be 105 grams, rather than 100 grams. So, you would calculate:

density = mass / volume
density = 105 grams / 100 ml
density = 1.05 g/ml

The effect on density would be that it would erroneously appear to be greater

Hope this helps!

Good luck

6 0

3 years ago

12. A beginner's bowling ball has a mass of 4.9 kg and a volume of 5.4 liters. Will it float in water,

Greeley [361]

Taking into account the definition of density and Archimedes' principle, the beginner bowling ball will float on the water.

But first it is neccesary to know that density is a quantity referred to the amount of mass in a certain volume of a substance or a solid object.

In other words, the density is the relationship between the weight (mass) of a substance and the volume that the same substance occupies.

The expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:

$density=\frac{mass}{volume}$

In this case, a beginner's bowling ball has a mass of 4.9 kg and a volume of 5.4 liters. This is:

mass= 4.9 kg= 4900 g (being 1 kg= 1000 kg)
volume= 5.4 L= 5400 mL (being 1L=1000 mL)

Replacing in the definition of density:

$density=\frac{4900 g}{5400 mL}$

Solving:

density=0.907 $\frac{g}{mL}$

On the other hand, Archimedes' principle says that an object immersed in a liquid experiences an upward vertical force equal to the weight of the volume of the dislodged liquid.

The sinking or floating of an object is determined by its density with respect to that of the liquid in which it is submerged.

Considering water as the liquid where the object is submerged in this case, an object with a higher density than water will sink. In contrast, an object with a lower density than water will float.

In this case, considering that water has a density of 1 $\frac{g}{mL}$ , the bowling ball for beginners has a lower density. This indicates that, having a lower density than water, the object will float.

In summary, the beginner bowling ball will float on the water.

Learn more about density:

5 0

3 years ago

How many moles of ions would you expect in an aqueous solution containing one mole of chromium(III) chloride? Hint: write out th

Cerrena [4.2K]

Answer:

Four

Explanation:

AlCl₃(aq) ⟶ Al³⁺(aq) + 3Cl⁻(aq)

One mole of AlCl₃ produces 1 mol of Al³⁺ and 3 mol of Cl⁻.

That's four moles of ions.

8 0

3 years ago

How many molecules of copper sulfate are needed to produce 5.0×1024 molecules of sodium sulfate?

lubasha [3.4K]

Answer:

5.0 × 10²⁴ molecules

Explanation:

Step 1: Write the balanced double displacement reaction

2 NaOH + CuSO₄ ⇒ Na₂SO₄ + Cu(OH)₂

Step 2: Calculate the moles corresponding to 5.0 × 10²⁴ molecules of Na₂SO₄

We will use Avogadro's number: there are 6.02 × 10²³ molecules in 1 mole of molecules.

5.0 × 10²⁴ molecule × 1 mol/6.02 × 10²³ molecule = 8.3 mol

Step 3: Calculate the moles of CuSO₄ required to produce 8.3 moles of Na₂SO₄

The molar ratio of CuSO₄ to Na₂SO₄ is 1:1. The moles of CuSO₄ required are 1/1 × 8.3 mol = 8.3 mol.

Step 4: Calculate the molecules corresponding to 8.3 moles of CuSO₄

We will use Avogadro's number.

8.3 mol × 6.02 × 10²³ molecule/1 mol = 5.0 × 10²⁴ molecule

4 0

3 years ago