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goldfiish [28.3K]
3 years ago
6

What causes your vocal cords to vibrate a)mucas b)air c)chemicals

Chemistry
1 answer:
Wewaii [24]3 years ago
5 0

Answer:

air

Explanation:

Vocal folds vibrate when excited by aerodynamic phenomena; they are not plucked like a guitar string. Air pressure from the lungs controls the open phase. The passing air column creates a trailing “Bernoulli effect,” which controls the close phase.

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A 20.0 mL solution of NaOH is neutralized with 24.1 mL of 0.200 M HBr. What is the concentration of the original NaOH solution
Alinara [238K]

Answer:

0.241 M

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

HBr + NaOH —> NaBr + H₂O

From the balanced equation above,

The mole ratio of acid, HBr (nₐ) = 1

The mole ratio of base, NaOH (n₆) = 1

Finally, we shall determine the concentration of the NaOH solution. This can be obtained as follow:

Volume of base, NaOH (V₆) = 20 mL

Volume of acid, HBr (Vₐ) = 24.1 mL

Concentration of acid, HBr (Cₐ) = 0.2 M

Concentration of base, NaOH (C₆) =?

CₐVₐ / C₆V₆ = nₐ/n₆

0.2 × 24.1 / C₆ × 20 = 1/1

4.82 / C₆ × 20 = 1

Cross multiply

C₆ × 20 = 4.82

Divide both side by 20

C₆ = 4.82 / 20

C₆ = 0.241 M

Therefore, the concentration of the NaOH solution is 0.241 M

8 0
3 years ago
For each equation , write all possible mole ratios . A. 2HgO(s)>2Hg(l)+O2(g) B. 4NH3(g)+6NO(g)>5N2(g)+6H2O(l)
Dafna11 [192]

A) For balanced chemical equation: 2HgO(s) → 2Hg(l) + O₂(g).

1) Mole ratio 1: n(HgO) : n(Hg) = 2 : 2 (1 : 1).

2) Mole ratio 2: n(HgO) : n(O₂) = 2 : 1.

3) Mole ratio 3: n(Hg) : n(O₂) = 2 : 1.

B) Balanced chemical equation: 4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l).

1) Mole ratio 1: n(NH₃) : n(NO) = 4 : 6 (2 : 3).

2) Mole ratio 2: n(NH₃) : n(N₂) = 4 : 5.

3) Mole ratio 3: n(NH₃) : n(H₂O) = 4 : 6 (2 : 3).

4) Mole ratio 4: n(NO) : n(N₂) = 6 : 5.

5) Mole ratio 5: n(NO) : n(H₂O) = 6 : 6 (1 :1).

6) Mole ratio 6: n(N₂) : n(H₂O) = 5 : 6.

3 0
3 years ago
Read 2 more answers
Consider a pot of water at 100 C. If it took 1,048,815 J of energy to vaporize the water and heat it to 135 C, how many grams of
jeka57 [31]

Answer:

There was 450.068g of water in the pot.

Explanation:

Latent heat of vaporisation = 2260 kJ/kg = 2260 J/g = L

Specific Heat of Steam = 2.010 kJ/kg C = 2.010 J/g = s

Let m = x g be the weight of water in the pot.

Energy required to vaporise water = mL = 2260x

Energy required to raise the temperature of water from 100 C to 135 C = msΔT = 70.35x

Total energy required = 2260x+x\times2.010\times(135-100)=2260x+70.35x=2330.35x

2330.35x=1048815\\x=450.068g

Hence, there was 450.068g of water in the pot.

8 0
2 years ago
Which word is the longest word in chemistry?​
Hunter-Best [27]
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4 0
2 years ago
Which of the following answers is true for the following statement?
Vsevolod [243]

it is either "aweak acid or a lousy (or very weak) acid"

6 0
2 years ago
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