How many moles of Fe2O3 are in 209 g of the compound?

Verizon [17]

1) Humans have how many sets of teeth?

C) 4

2) Adults have how may permanent teeth?

B) 32

3) What type of food products make teeth stronger?

Cheese and Yogurt

4) What is the visible part of the tooth?

C) crown

5) What part of the tooth is harder than bone?

D) dentin

6) What part of the tooth is soft and contains the nerves?
A) pulp

7) What refers to the alignment of the teeth?

Occlusion

8) The lower teeth protect what?

D) tongue

9) Malocclusion is the least common reason for referral to orthodontists.

False

10) Usually, malocclusion is hereditary.
True

11) Which type of malocclusion is commonly referred to as an overbite?

Class Two

12) Bacteria are living organisms than cause cavities.
True

13) What is the whitish film that grows on your teeth called?

Plaque

14) What is the best fighter of cavities?

Fluoride

Gingivitis is the inflammation of the
A) gums

16) Gingivitis is permanent.

False

17) Name 2 ways to fight gingivitis.

Proper Toothbrushing and Flossing
Professional Dental Cleanings

18) The cause of gum disease is

B) bacteria

19) Untreated, periodontitis will progress until your teeth loosen and fall out.
True

6 0

3 years ago

Write the chemical formula for the cation present in the aqueous solution of cuso4.

Inessa05 [86]

The ionic compounds on dissolving in aqueous solution dissociate into their respective ions that are cations and anions.

The ions present in the aqueous solution of $CuSO_4$ are $Cu^{2+}$ and $SO_{4}^{2-}$ .

So, the chemical formula of the cation present in the aqueous solution of $CuSO_4$ is $Cu^{2+}$ .

4 0

3 years ago

Mercury is a liquid metal having a density of 13.6 gml what is the volume of 1.00 lb of mercury metal

tresset_1 [31]

The volume of 1.00 lb of mercury metal : 33.352 ml

<h3>Further explanation </h3>

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

Density formula:

$\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}$

ρ = density

m = mass

v = volume

Density of metal(mercury) = 13.6 g/ml

mass of mercury : 1 lb = 453,592 g

So the volume :

$\tt V=\dfrac{453,592~g}{13.6~g/ml}\\\\V=33.352~ml$

7 0

3 years ago

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. what volume of hydrogen chloride

Anna71 [15]

The reaction between methane gas and chlorine gas to form hydrogen chloride and carbon tetrachloride, all in their gaseous form can be expressed through the chemical reaction below.

CH₄ + 4Cl₂ --> 4HCl + CCl₄

Let us assume that all the involved gases behaves ideally such that each mole of the gas is equal to 22.4 L.

Through proper dimensional analysis, the volume of the produced hydrogen chloride is calculated,
V(HCl) = (1.69 mL CH₄)(1 L CH₄/ 1000 mL CH₄)(1 mol CH₄/22.4 L CH₄)(4 mols HCl/1 mol CH₄)(22.4 L HCl/1 mol HCl)(1000 mL/1 L)

V(HCl) = 6.76 mL

ANSWER: 6.76 mL

3 0

3 years ago

Do parts a, b and c

Leona [35]

Answer:- (a)The pH of the buffer solution is 3.90.

(b) the pH of the solution after addition of HCl would be 3.60.

(c) the pH of the buffer solution after addition of NaOH is 4.32.

Solution:- (a) It is a buffer solution so the pH could easily be calculated using Handerson equation:

$pH=Pka+log(\frac{base}{acid})$

pKa can be calculated from given Ka value as:

$pKa=-logKa$

$pKa=-log(6.3*10^-^5)$

pKa = 4.20

let's plug in the values in the Handerson equation:

$pH=4.20+log(\frac{0.025}{0.05})$

pH = 4.20 - 0.30

pH = 3.90

The pH of the buffer solution is 3.90.

(b) Let's say the acid is represented by HA and the base is represented by $A^-$ .

Original mili moles of HA from part a = 0.05(100) = 5

original mili moles of $A^-$ from part a = 0.025(100) = 2.5

mili moles of HCl that is $H^+$ added = 0.100(10.0) = 1

This HCl reacts with the base present in the buffer to make HA as:

$A^-+H^+\rightarrow HA$

Total mili moles of HA after addition of HCl = 5+1 = 6

mili moles of base after addition of HCl = 2.5-1 = 1.5

Let's plug in the values in the Handerson equation again. Here, we could use the mili moles to calculate the pH. The answer remains same even if we use the concentrations also as the final volume is same both for acid and base.

$pH=4.20+log(\frac{1.5}{6})$