Tidal power plants work best in small estuaries or intels that border the ocean. True or false?

A student must use 220 mL of hot water in a lab procedure. Calculate the amount of heat required to raise the temperature of 220

Inessa [10]

Answer:

64,433.6 Joules

Explanation:

<u>We are given</u>;

Volume of water as 220 mL
Initial temperature as 30°C
Final temperature as 100°C
Specific heat capacity of water as 4.184 J/g°C

We are required to calculate the amount of heat required to raise the temperature.

We know that amount of heat is calculated by;

Q = mcΔT , where m is the mass, c is the specific heat, ΔT is the change in temperature.

Density of water is 1 g/mL

Thus, mass of water is 220 g

ΔT = 100°C - 30°C

= 70°C

Therefore;

Amount of heat, Q = 220g × 4.184 J/g°C × 70°C

= 64,433.6 Joules

Thus, the amount of heat required to raise the temperature of water is 64,433.6 Joules

7 0

3 years ago

Both protons and neutrons are located inside the nucleus of an atom. Changing the number of neutrons in the nucleus would form d

skelet666 [1.2K]

By changing the number of protons in an element, you are fundamentally changing it to a different element.

It’s worth remembering that:

Protons identify the element

Electrons determine properties and behavior of the element

Neutrons contribute to isotopes and just help keep everything stable, by separating the charges. Remember opposites attract. Like charges repel.

6 0

3 years ago

A student determines the aluminum content of a solution by first precipitating it as aluminum hydroxide, and then decomposing th

Vladimir [108]

Answer: The student should obtain <u>1.103 g of aluminum oxide </u>

Explanation:

First we write down the equations that represent the aluminum hydroxide precipitation from the reaction between the aluminum nitrate and the sodium hydroxide:

Al(NO3)3 + 3NaOH → 3NaNO3 + Al(OH)3

Now, the equation that represents the decomposition of the hydroxide to aluminum oxide by heating it.

2Al(OH)3 → Al2O3 + 3H2O

Second, we gather the information what we are going to use in our calculations.

Volumen of Al(NO3)3 = 40mL

Molar concentration of Al(NO3)3 = 0.541M

Molecular Weight Al2O3 = 101.96 g/mol

Third, we start using the molar concentration of the aluminum nitrate and volume used to find out the total amount of moles that are reacting

$\frac{0.541moles Al(NO3)3}{1L} x\frac{1L}{1000mL} x 40mL Al(NO3)3 = 0.022moles Al(NO3)3$

then we use the molar coefficients from the equations to discover the amount of Al2O3 moles produced

$0.022moles Al(NO3)3 x \frac{1mol Al(OH)3}{1mol Al(NO3)3} X\frac{1mol Al2O3}{2molAl(OH)3} = 0.011 moles Al2O3$

finally, we use the molecular weight of the Al2O3 to calculate the final mass produced.

$0.011moles Al2O3 x \frac{101.96g Al2O3}{1mol Al2O3} = 1.103g Al2O3$

4 0

4 years ago

Right balanced chemical equations, complete ionic equations, and net ionic equations for the reactants below. Include a descript

Ronch [10]

I haven't learned this yet but my friend did I'll text them and get back to you with the answer in a bit brb!

5 0

3 years ago

The yearly amounts of carbon emissions from cars in Belgium are normally distributed with a mean of 13.9 gigagrams per year and

zavuch27 [327]

Answer:

The correct option is;

c. 0.167

Explanation:

The parameters given are;

The mean, μ = 13.9 Gigagrams/year

The standard deviation, σ = 5.8 Gigagrams/year

The z-score formula is given as follows;

$z = \dfrac{x - \mu }{\sigma }$

Where:

x = Observed score = 11.5 Gigagrams/year

We have;

$z = \dfrac{11.5 - 13.9 }{5.8 } = \dfrac{-2.4 }{5.8 } = 0.4138$

From the z-score table relations/computation, the probability (p-value) = 0.6605

Where:

x = Observed score = 14 Gigagrams/year

We have;

$z = \dfrac{14- 13.9 }{5.8 } = \dfrac{0.1}{5.8 } = 0.01724$

From the z-score table relations/computation, the probability (p-value) = 0.4931

Therefore, the probability, $p_{ca}$ , that the amount of carbon emissions from cars in Belgium for a randomly selected year are between 15.5 Gigagrams/year and 14 Gigagrams/year = The area under the normal curve bounded by the p-values for the two amounts of carbon emission

Which gives;

$p_{ca}$ = 0.6605 - 0.4931 = 0.1674 ≈ 0.167

Therefore, the correct option is c. 0.167.

4 0

4 years ago