Problem #1 I don’t understand this stuff

To aid in the prevention of tooth decay, it is recommended that drinking water contains 0.900 ppm fluoride (F-). A) How many g o

Romashka-Z-Leto [24]

Answer:

a) 1.740 g of F- must be added to a cylindrical water reservoir

b) Grams of sodium fluoride, NaF, that contain this much fluoride:

3.84 g

Explanation:

Step 1. calculate the volume of the tank:

Volume of cylinder =

$\pi r^{2}h$ ,

Here r = radius of the cylinder = d/2

h = depth = 21.80m

$r=\frac{d}{2}$

$=\frac{3.36x10^{2}}{2}$

= 168 m

Volume =

$=\frac{22\times 168^{2}\times 21.80}{7}$

$=1.93\times 10^{6} m^{3}$

2.Convert ppm to g/m3 and Solve for mass of F-

$1ppm = 1g/m^{3}$

$0.9ppm = 0.9g/m^{3}$

Because both ppm and g/m3 are same quantity .

$g/m^{3} =\frac{mass\ of\ F-(g)}{Volume\ m^{3}}\times 10^{6}$

$0.9 =\frac{mass\ of\ F-}{1.93\times 10^{6} m^{3}}\times 10^{6}$

$mass\ of\ F- =1.740g$

mass of F- required = 1.740 g

3. Apply mole concept to calculate grams of sodium fluoride produced

mass of 1 mole of F2 = 38 g

mass of 1 mole of NaF = 42 g

(from periodic table calculate molar mass)

$2Na+F_{2}\rightarrow 2NaF$

Here 1 mole of F2 produce = 2 mole of NaF

So,

38 g of F2 produce = 2 x 42 g of NaF

38 g of F2 produce = 84 g of NaF

1 g of F2 produce = 84/38 g of NaF

1.74 g F2 produce =

$\frac {84}{38}\times 1.74$

1.74 g F2 produce = 3.84 g of NaF

3.84 g of NaF is produced

3 0

3 years ago

What do budding, binary fishing and spore formation have in commen?

vladimir2022 [97]

Answer:

all are forms of asexual reproduction

3 0

3 years ago

If 15.0 liters of neon at 25.0 °C is allowed to expand to 45.0 liters, what must the new temperature be to maintain constant pre

Fed [463]

Answer:

893K=620 °C

Explanation:

PV=nRT

P=nRT/V

P= (1 mol* 0.0821* 298K)/15L P=1.63 atm

T=PV/nR

T=(1.63 atm*45 L)/(1 mol*0.0821)

T= 893K=620 °C

7 0

3 years ago

Rank the following chemical species from largest ionic radius to smallest ionic radius

7nadin3 [17]

Answer:

Biggest Radii V²⁺ > V³⁺ > V⁴⁺ > V⁵⁺ Smallest Radii

General Formulas and Concepts:

Periodic Trends: Atomic/Ionic Radii
Coulomb's Law

Explanation:

The Periodic Trend for Atomic Radii is down and to the left. Therefore, the element with the largest radius would be in the bottom left corner of the Periodic Table.

Anions will always have a bigger radii than the parent radii. When we add e⁻ to the element, we are increasing the e⁻/e⁻ repulsions. This will cause e⁻ to repel themselves more and thus create more space, increasing the radii size.

Cations will always have smaller radii than the parent radii. When we remove e⁻ from the element, we are decreasing e⁻/e⁻ repulsions. Since there are less e⁻, there is no need for more space and thus decreases the radii size.

Since Cations are smaller than the parent radii, the more e⁻ we remove, the smaller it will become.

Therefore, the least removed e⁻ Vanadium would be the largest and the most removed e⁻ Vanadium would be the smallest.

4 0

3 years ago

What is the abbreviation for water?

DENIUS [597]

Abbreviation is: H2O

5 0

3 years ago

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