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Verdich [7]
3 years ago
11

There are two parts to a wave, the

Chemistry
1 answer:
Tema [17]3 years ago
8 0
Start with the process of elimination. Obviously, color is not a specific component to a wave, so we can cancel that out. While there are frequencies and measurable speeds of waves, those are not considered parts of the wave; nor is the wavelength or the base considered to be a part of the wave. The crest is the highest point of the wave, and is considered a part of the wave, as well as the trough, which is the the lowest point on the wave.
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Sucrose is very soluble in water. at 25◦c, 211.4 grams of sucrose will dissolve in 100 g of water. given that the density of the
stiv31 [10]
Molarity of solution is mathematically expressed as,
M = \frac{x\text{weight of solute(g)}}{\text{Molecular weight X Volume of solution(l)}}

We know that volume = mass/density
Given: mass of solution = 100 g, Density = 1.34 g/ml
∴ volume = 100/1.34 = 88.49 ml = 0.08849 l

Also, we know that molecular weight of sucrose = 342.3 g/mol
 ∴M = \frac{x\text{211.4}}{\text{342.3 X 0.08849}}
       = 6.979 M

Thus, molarity of solution is 6.979 M
8 0
3 years ago
An experiment requires that enough C3H8O be used to yield of oxygen .
blsea [12.9K]

Answer: So if you had 570 cm of ribbon, then 570%2F8.5=67.05 which means that about 67 students can do the experiment (round down to the nearest whole number).

Explanation: If you had 8.5 cm of ribbon, then only 8.5%2F8.5=1 student can do the experiment. If you had 17 cm of ribbon, then 17%2F8.5=2 students can do the experiment.

8 0
3 years ago
2. A solution is made by dissolving 3.88 g of NaCl in enough water to make 67.8 mL of solution. What is the concentration of sod
Sauron [17]

Answer:

is 23

Explanation:

i got it rigjt .......

3 0
2 years ago
Convert the following measurements using dimensional analysis. Set up problem using factors. Cross out the units that cancel. YO
uranmaximum [27]

Answer:

1) 0.423 m

2) 3.107 mi

3) 68.18 kg

4) 0.0083 mem

5) 0.528 gal

6) 4300 mL

7) 32.4 mem

8) 523.013 km

9) 70.866 in

10) 2.3 yek

Note: I can’t type the about equal to sign or the sign that shows a repeating decimal, so check the image for that and my work.

Explanation:

4 0
3 years ago
7.00 of Compound x with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25c.
beks73 [17]

Answer:

\Delta H_{f,C_3H_4}=276.8kJ/mol

Explanation:

Hello!

In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

\Delta H_{rxn} =- m_wC_w\Delta T

We plug in the mass of water, temperature change and specific heat to obtain:

\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ

Now, this enthalpy of reaction corresponds to the combustion of propyne:

C_3H_4+4O_2\rightarrow 3CO_2+2H_2O

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol

Now, we solve for the enthalpy of formation of C3H4 as shown below:

\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol

Best regards!

7 0
3 years ago
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