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Kryger [21]
3 years ago
6

What happens if your caculating the speed of a longitudinal wave and you increase the temp in the room

Chemistry
1 answer:
nordsb [41]3 years ago
5 0

The speed of a longitudinal wave would increase, as the kinetic energy molecules would move quicker in response to the higher temperature.

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AURORKA [14]
Can u write that in English
6 0
4 years ago
You titrate 25.00 mL of the lemon-lime Kool-Aid (with KI, HCl, and starch) with 0.001000 M KIO3(aq) solution. The titration requ
Svetradugi [14.3K]

Answer:

1.0190 x 10⁻⁵ mol

Explanation:

We know the titration required 10.19 mL of 0.001000 M KIO₃, from this information we can calculate the number of moles KIO₃ reacted and from there the number of moles of ascorbic acid since it is a monoprotic acid ( 1 equivalent of ascorbic acid to one equivalent KIO₃).

Molarity = mol/V

V KIO₃ = 10.19 mL = 10.19 mL x 1 L/1000 mL = 0.01019 L

⇒ mol KIO₃ = V x M = 0.01019 L x 0.0010 mol / L =  1.0190 x 10⁻⁵ mol KIO₃

# mol ascorbic acid = # mol KIO₃ = 1.0190 x 10⁻⁵ mol

8 0
3 years ago
Rust forming on a piece of iron is an example of which of the following? *
kvasek [131]
Rust is an iron oxide and formed by the reaction of iron and oxygen in the presence of moisture. So the answer would be C
8 0
3 years ago
Given the amount of camphor (200mg) we are using in this experiment, please determine how many mg of sodium borohydride to use i
swat32

Answer:

Explanation:

From the given information:

Camphor may be reduced as readily in the presence of sodium borohydride(NaHB4). The resulting compound which is stereoselective requires 1 mole of sodium borohydride (NaHB4) to reduce 1 mole of camphor in this reaction. The reaction is shown below.

Through the reduction process of camphor, the reducing agent can reach the carbonyl face with a one-carbon linkage. The product stereoisomer is known as borneol.

If the molecular weight of camphor = 152.24 g/mol

and it mass = 200 mg

The its no of moles = 200 mg/ 152.24 g/mol

= 1.3137 mmol

Now the amount of the required mmol for NaBH4 to be consumed in the reaction = 5.2 × 1.3137 mmol

= 6.831 mmol

since the molar mass of NaBH4 = 37.83 g/mol

Then, using the same formula:

No of moles = mass/molar mass

mass = No of moles × molar mass

mass = 6.831 mmol × 37.83 g/mol

mass of NaBH4 used = 258.42 mg  

7 0
3 years ago
Potassium iodide reacts with lead(II) nitrate in this precipitation reaction: 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s) Wha
andrew11 [14]

Answer:

a. 174 mL

Explanation:

Let's consider the following reaction.

2 KI(aq) + Pb(NO₃)₂(aq) → 2 KNO₃(aq) + PbI₂(s)

We have 155.0 mL of a 0.112 M lead(II) nitrate solution. The moles of Pb(NO₃)₂ are:

0.1550 L × 0.112 mol/L = 0.0174 mol

The molar ratio of KI to Pb(NO₃)₂ is 2:1. The moles of KI are:

2 × 0.0174 mol = 0.0348 mol

The volume of a 0.200 M KI solution that contains 0.0348 moles is:

0.0348 mol × (1 L / 0.200 mol) = 0.174 L = 174 mL

5 0
3 years ago
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