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olga2289 [7]
3 years ago
12

Electromagnetic vs. Mechanical Waves 1?

Chemistry
1 answer:
Tomtit [17]3 years ago
6 0

Answer:

example of electromagnetic waves water waves and sound wave.

Types of mechanical waves are transverse and longitudinal wave...

You might be interested in
potassium, k, has an atomic number of 19 and an atomic mass of 39 how many electrons does a neutral atom of K have?
coldgirl [10]
A neutral atom of potassium has 19 electrons.
8 0
4 years ago
Compute the values. Express these answers to the hundredths place (i.e., two digits after the decimal point). log ( 2.1 ) = ln (
zepelin [54]

Answer:

x = 4.17

y = 1.86

Explanation:

0.62 = log(x)

x = 10^0.62 = 4.17 ( to the nearest hundredth)

0.62 = ln(y)

y = e^0.62 = 1.86 (to the nearest hundredth)

8 0
4 years ago
If 2 CH3OH + 3 O2 -> 2CO2 + 4 H2O was carried out in the laboratory and 219 g of water was produced, what would the percent y
ANEK [815]

Answer:

Percent yield = 84.5 %

Explanation:

Given data:

Mass of methanol = 229 g

Actual yield of water = 219 g

Percent yield of water = ?

Solution:

Chemical equation:

2CH₃OH + 3O₂  →  2CO₂  + 4H₂O

Number of moles of methanol:

Number of moles = mass/ molar mass

Number of moles = 229 g/ 32 g/mol

Number of moles = 7.2 mol

Now we will compare the moles of water with methanol.

                        CH₃OH         :            H₂O

                            2               :               4

                           7.2             :           4/2×7.2 = 14.4 mol

Mass of water:

Mass = number of moles × molae mass

Mass = 14.4 mol × 18 g/mol

Mass = 259.2 g

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 219 g / 259.2 g × 100

Percent yield = 84.5 %

7 0
3 years ago
Balance each skeleton reaction, use Appendix D to calculate E°cell , and state whether the reaction is spontaneous:(a) Cu⁺(aq) +
melamori03 [73]

The balanced equation reaction :

2Cu+(aq) + PbO2(s) + SO42–(aq) + 4H+  -----------> PbSO4(s) + 2 Cu2+(aq)  + 2H2O

A balanced equation is an equation for a chemical reaction in which the range of atoms for every detail in the reaction and the full charge is identical for both the reactants and the products. In other phrases, the mass and the fee are balanced in each aspect of the reaction.

A balanced equation happens while the variety of the atoms involved inside the reactants side is identical to the variety of atoms in the goods facet. in this chemical reaction, nitrogen (N2) reacts with hydrogen (H) to provide ammonia (NH3). The reactants are nitrogen and hydrogen, and the product is ammonia.

A chemical equation is the symbolic representation of a chemical reaction inside the form of symbols and formulae, wherein the reactant entities are given on the left-hand side and the product entities.

Eo  = Eored - Eo oxd

 = 1.69  - 0.153

 = 1.54 V

Eocell is positive so the reaction is spontaneous.

Learn more about the balanced reaction  here brainly.com/question/15355912

#SPJ4

4 0
2 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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