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ziro4ka [17]
3 years ago
9

Which of the following microscopic soil life forms is also known as a “ring worm”?

Chemistry
2 answers:
UNO [17]3 years ago
8 0
Protozoa is known as ring worm
OverLord2011 [107]3 years ago
3 0

Answer:

nematodes

Explanation:

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In a combustion reaction, ethanol (C2H5OH) is burned in the presence of oxygen (O2), producing carbon dioxide (CO2) and water (H
charle [14.2K]
Answer:
Reactants: Ethanol (C₅H₅OH), Oxygen (O₂)

Products: Carbon dioxide (CO₂) and Water (H₂O)

Explanation:
Let’s rewrite each part of the described situation in chemical equation terms.

Ethanol - C₂H₅OH
is burned in the prescence of - +
oxygen - O₂
producing - →
carbon dioxide - CO₂
and - +
water - H₂O

Chemical equation:
C₂H₅OH + O₂ → CO₂ + H₂O

The terms on the left side of a chemical equation are reactants, and the terms on the right side of a chemical equation are products.

Therefore, since we know this, ethanol (C₂H₅OH) & oxygen (O₂) are the reactants, and carbon dioxide (CO₂) & water (H₂O) are the products.
8 0
3 years ago
The extent of ionization of a weak electrolyte is increased by adding to the solution a strong electrolyte that has an ion in co
Anvisha [2.4K]
It’s false!




Hope this helps:)


Explanation:

7 0
3 years ago
A 24.00 mL sample of a solution of Pb(ClO3)2 was diluted with water to 52.00 mL. A 17.00 mL sample of the dilute solution was fo
klio [65]

Answer:

0.238 M

Explanation:

A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.

C₁.V₁ = C₂.V₂

C₁ × 24.00 mL = 0.220 M × 52.00 mL

C₁ = 0.477 M

The concentration of Pb(ClO₃)₂ is:

\frac{0.477molClO_{3}^{-} }{L} \times \frac{1molPb(ClO_{3})_{2}}{2molClO_{3}^{-}} =0.238M

4 0
3 years ago
The combustion of 1.685 g of propanol (C3H7OH) increases the temperature of a bomb calorimeter from 298.00 K to 302.16 K. The he
AfilCa [17]

Answer:

ΔH =  - 2020.57 kJ/mol

Explanation:

Given that :

mass of propanol = 1.685 g

the molar molar mass = 60 g/mol

Thus; the number of  moles = mass/molar mass

= 1.685 g/60 g/mol

= 0.028 g/mol

However ;

ΔH = heat capacity C × Δ T

Given that:

The temperature increases from  298.00 K to 302.16 K.

Then ;

Δ T = 302.16 K - 298.00 K

Δ T = 4.16 K

heat capacity C = 13.60 kJ/K

∴

ΔH = 13.60 kJ/K × 4.16 K

ΔH =  56.576 kJ

The equation of the given reaction can be represented as :

C_3H_7OH_{(l)}+\dfrac{3}{2}O_{2(g)}  \to 3CO_{2(g)} +4H_2O_{(l)}

Thus for 0.028 mol of heat liberated; ΔH =  56.576 kJ

For 1 mole of heat liberated now:

ΔH =  56.576 kJ/0.028 mol

ΔH =  2020.57 kJ/mol

SInce , Heat is liberated, the reaction undergoes an exothermic reaction thus;

ΔH =  - 2020.57 kJ/mol

5 0
3 years ago
Read 2 more answers
What is the ph of 0.450 m al(no3)3 [ka for al3+(aq) = 1.00x10-5]? express your answer to two decimal places?
White raven [17]

Answer : The pH of the solution is, 2.67

Explanation :

The equilibrium chemical reaction is:

                           Al^{3+}+H_2O\rightarrow Al(OH)^{2+}+H^+

Initial conc.       0.450                   0               0

At eqm.           (0.450-x)                 x               x

As we are given:

K_a=1.00\times 10^{-5}

The expression for equilibrium constant is:

K_a=\frac{(x)\times (x)}{(0.450-x)}

Now put all the given values in this expression, we get:

1.00\times 10^{-5}=\frac{(x)\times (x)}{(0.450-x)}

x=0.00212M

The concentration of H^+ = x = 0.00212 M

Now we have to calculate the pH of solution.

pH=-\log [H^+]

pH=-\log (0.00212)

pH=2.67

Therefore, the pH of the solution is, 2.67

8 0
3 years ago
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