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Oxana [17]
3 years ago
13

What is a wind turbine often used for?

Chemistry
2 answers:
kirill115 [55]3 years ago
5 0

Answer:

A

Explanation:

Converting thermal energy into electrical energy

Colt1911 [192]3 years ago
3 0

Wind turbines are used for converting the energy in sunlight into mechanical energy. It is D.

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What is the first step in the scientific method? A. form a hypothesis B. illustrate the process C. observe and ask questions D.
Tom [10]
The answer would be C) Observe and ask questions because if we do process of elimination we see that drawing conclusions is NOT the first step, forming a hypothesis and illustrating the process is part of the method, but NOT the first step.
3 0
3 years ago
Read 2 more answers
A volume of 129 mL of hydrogen is collected over water. The water level in the collecting vessel is the same as the outside leve
mixas84 [53]

Explanation:

As it is given that water level is same as outside which means that theoretically, P = 756.0 torr.

So, using ideal gas equation we will calculate the number of moles as follows.

                  PV = nRT

or,           n = \frac{PV}{RT}

                 = \frac{\frac{756}{760}atm \times 0.129 L}{0.0821 Latm/mol K \times 298 K}

                  = 0.0052 mol

Also,  No. of moles = \frac{mass}{\text{molar mass}}

               0.0052 mol = \frac{mass}{2 g/mol}

                  mass = 0.0104 g

As some of the water over which the hydrogen gas has been collected is present in the form of water vapor. Therefore, at 25^{o}C

                P_{\text{water vapor}} = 24 mm Hg

                                = \frac{24}{760} atm

                                = 0.03158 atm

Now,   P = \frac{756}{760} - 0.03158

              = 0.963 atm

Hence,   n = \frac{0.963 atm \times 0.129 L}{0.0821 L atm/mol K \times 298 K}

                 = 0.0056 mol

So, mass of H_{2} = 0.0056 mol × 2

                         = 0.01013 g (actual yield)

Therefore, calculate the percentage yield as follows.

      Percent yield = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100

                              = \frac{0.01013 g}{0.0104 g} \times 100            

                              = 97.49%

Thus, we can conclude that the percent yield of hydrogen for the given reaction is 97.49%.

3 0
3 years ago
What kind of reaction is the equation?<br> Zn(s) + 2HCl(aq) → ZnCl2 (aq) + H2(g) + heat
QveST [7]
Given the balanced equation representing a reaction: Zn(s) + 2HCl(aq) + H2(g) + ZnCl2(aq) Which statement is true about energy in this reaction? A. The reaction is exothermic because it releases heat.
3 0
2 years ago
A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M Na
s344n2d4d5 [400]

Answer:

a: before equivalence point

b: equivalence point

c: before equivalence point

d: after the eqivalence point

e: before equivalence point

f:  after the eqivalence point

Explanation:

Balanced equation of reaction:

NaOH +HCl =NaCl +H2O;

Volume of HCl is fixed and it 100ml and concentration is 1.0M

N1 and N2 normality of HCl and NaOH respectively;

V1 and V2 volume of HCl and NaOH respectively;

we have given molarity but we need normality;

Normality=molarity \times n-factor

<em>but in case of NaOH and HCl n-factor is 1 for each.</em>

hence

normality=molarity;

At equivalence point:  N_1V_1=N_2V_2

Before equivalence point : N_1V_1>N_2V_2

After the equivalence point: N_1V_1

N_1V_1=100\times1=100

case a:  5.00 mL of 1.00 M NaOH

N_2V_2=5\times1=5

N_1V_1>N_2V_2 hence it is before equivalence point

case b: 100mL of 1.00 M NaOH

N_2V_2=100\times1=100

N_1V_1=N_2V_2 hence it is equivalence point

case c:  10.0 mL of 1.00 M NaOH

N_2V_2=10\times1=10

N_1V_1>N_2V_2 hence it is before equivalence point

case d: 150 mL of 1.00 M NaOH

N_2V_2=150\times1=150

N_1V_1 hence it is after the eqivalence point

case e: 50.0 mL of 1.00 M NaOH

N_2V_2=50\times1=50

N_1V_1>N_2V_2 hence it is before equivalence point

case f: 200 mL of 1.00 M NaOH

N_2V_2=200\times1=200

N_1V_1 hence it is after the eqivalence point

7 0
3 years ago
If the actual yield of sodium chloride from the reaction of
lions [1.4K]

If the actual yield of sodium chloride from the reaction of 8.3 g of sodium and 4.5 g of chlorine is 6.4 g, what is the percent yield?

 

Answer : 2Na(s) + Cl2(g) → 2NaCl(s)

Explanation: Friend me!!!

7 0
2 years ago
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