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tankabanditka [31]
2 years ago
5

Why are fungi classified together?

Chemistry
1 answer:
mash [69]2 years ago
4 0

Answer:

A, they get their food in the same way.

Explanation:

Some fungi can be toxic or poisonous.

Example: mushrooms. Mushrooms typically are found in forests and meadows/plains.

Some fungi are glowing, brown, and in many forms/shapes.

In conclusion, fungi all get nutrient from the dirt they grew in, forests can be moist which gives fungi excellent places to form, usually around trees, because they can absorb plenty of nutrients from the dirt.

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Calculate the mass of NaF that must be added to 300ml if a 0.25m HF solution to form a buffer solution with a ph of 3.5​
Kamila [148]

Answer:

6.574 g NaF into 300ml (0.25M HF) => Bfr with pH ~3.5

Explanation:

For buffer solution to have a pH-value of 3.5 the hydronium ion concentration [H⁺] must be 3.16 x 10⁻⁴M ( => [H⁺] = 10^-pH = 10⁻³°⁵ =3.16 x 10⁻⁴M).

Addition of NaF to 300ml of 0.25M HF gives a buffer solution. To determine mass of NaF needed use common ion analysis for HF/NaF and calculate molarity of NaF, then moles in 300ml the x formula wt => mass needed for 3.5 pH.

HF ⇄ H⁺ + F⁻; Ka = 6.6 x 10⁻⁴

Ka = [H⁺][F⁻]/[HF] = 6.6 x 10⁻⁴ = (3.16 x 10⁻⁴)[F⁻]/0.25 => [F⁻] = (6.6 x 10⁻⁴)(0.25)/(3.16x10⁻⁴) = 5.218M in F⁻ needed ( = NaF needed).

For the 300ml buffer solution, moles of NaF needed = Molarity x Volume(L)

= (5.218M)(0.300L) = 0.157 mole NaF needed x 42 g/mole = 6.574 g NaF needed.

Check using the Henderson - Hasselbalch Equation...

pH = pKa + log ([Base]/[Acid]); pKa (HF) = 3.18

Molarity of NaF = (6.572g/42g/mole)/(0.300 L soln) = 0.572M in NaF = 0.572M in F⁻.

pH = 3.18 + log ([0.572]/[0.25]) ≅ 3.5.

One can also back calculate through the Henderson -Hasselbalch Equation to determine base concentration, moles NaF then grams NaF.

4 0
3 years ago
A toothpaste contains sodium fluoride (NaF) What percentage of Fluoride is present.(4cs)
MariettaO [177]

Answer:

45.2%

Explanation:

To calculate the percent of an element in a compound we divide the molar mass of the element by the compound and multiply that by 100

First lets find the molar mass of Fluoride

Looking at the periodic table Fluoride has a molecular mass of 18.998 g

Now we need to find the molecular mass of NaF

Looking at a periodic table, Sodium (Na) has a molecular mass of 22.990g and Fluoride has a molecular mass of 18.998 so NaF has a molecular mass of 22.990(1) + 18.998(1) = 41.988g

Now we divide the mass of fluoride by the mass of sodium fluoride and multiply that by 100 to find the percentage of fluoride that is present in NaF

Mass of Fluoride = 18.998g

Mass of Sodium Fluoride = 41.988g

Percentage of fluoride present in NaF = (18.998g / 41.988g) * 100 = 45.2%

7 0
2 years ago
What type of chemical reaction is AgNO3(aq)+KCL(aq) AgCL(s)+KNO3(aq)​
emmasim [6.3K]

Answer:

double replacement reaction or double displacement reaction,

Explanation:

double replacement reaction, double displacement reaction, is a chemical process involving the exchange of bonds between two non-reacting chemical species which results in the creation of products with similar or identical bonding affiliations

Classically, these reactions result in the precipitation of one product.

in thía case it is AgCl

3 0
3 years ago
Write a balanced chemical equation to show the reaction of naoh with the monoprotic acid hcl.
Yakvenalex [24]
NaOH or sodium hydroxide is a base which reacts with HCl or Hydrochloric acid to from a salt and water. The equation that represents this reaction is as follows: NaOH (I) + HCl (l) ==H2O (l) +NaCl (aq). The reaction between an acid and a base always produces a salt and water. In this case the salt is sodium chloride. 
4 0
3 years ago
Calculate the equilibrium constant for the reaction using the balanced chemical equation and the concentrations of the substance
Oliga [24]

Answer:

4.75 is the equilibrium constant for the reaction.

Explanation:

CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)

Equilibrium concentration of reactants :

[CO]=0.0590 M,[H_2O]=0.00600 M

Equilibrium concentration of products:

[CO_2]=0.0410 M,[H_2]=0.0410 M

The expression of an equilibrium constant is given by :

K_c=\frac{[CO_2][H_2]}{[CO][H_2O]}

K_c=\frac{0.0410 M\times 0.0410 M}{0.0590M\times 0.00600 M}

K_c=4.75

4.75 is the equilibrium constant for the reaction.

3 0
2 years ago
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