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3 years ago

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We will begin this topic by practicing the use of the scientific method. The steps of the scientific method are: observation, hy

pothesis, experiment, results, and theory. Using your understanding of the scientific method from your readings and lessons, you will be applying this knowledge to a real world situation. Take a real world scenario in the news, or another outside source and apply the scientific method, being sure to detail the controls on your experiment. Describe a result that would confirm your hypothesis.

1 answer:

ludmilkaskok [199]3 years ago

5 0

Answer

i like it when kids rub my legs

Explanation:

bernie sanders

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When 20.4g of sodium metal are mixed with chlorine gas, are 52.0 g of sodium chloride

vagabundo [1.1K]

Answer: True

Explanation:

$\mathrm{The\ balanced \ chemical \ equation\ is\ :}\\$2 \mathrm{Na}{(s)}+\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{NaCl}(s)$

<u>For Na</u>

$\\$Given mass $=20.4 \mathrm{~g}$ \ \ Molecular mass $=23 \mathrm{~g}\ \ $Number of moles $=\frac{\text { Given mass }}{\text { Molecular mass }}$$$\begin{aligned}&=\frac{20.4}{23} \\&=0.887 \text { moles }\end{aligned}$$Stoichiometric ratio of $\mathrm{Na}$ is same as that of $\mathrm{NaCl}$ i.e., $1: 1$.$\therefore$ Moles of $\mathrm{NaCl}=0.887$ motes$

<u>For NaCl</u>

$\\\begin{aligned}\text { Molecular mass } &=23+35.5 \\&=58.5 \mathrm{~g}\end{aligned}$

$\begin{aligned}\\& $\therefore$ Mass of $\mathrm{NaCl}$ produced will be=\text { No.of moles } \times \text { Molecular mass } \\&=0.887 \times 58.5 \\&=51.9 \\&\simeq 52 \mathrm{~g} \quad\end{aligned}$$$

Therefore, it is true that when 20.4g of sodium metal are mixed with chlorine gas, 52.0 g of sodium chloride is produced

6 0

2 years ago

Read 2 more answers

Heat trapped from hot rock within Earth’s crust provides energy for electricity.

MaRussiya [10]

Answer:

IM NOT SURE!!! but this could be true

Explanation:

5 0

3 years ago

Aluminum oxide (Al2O3) occurs in nature as a mineral called corundum, which is noted for its hardness and resis- tance to attack

Neko [114]

Answer:

7.03×10²³ atoms of Al

Explanation:

Corundum density = 3.97g/cm³

Corundum = Al₂O₃. Then, you have 2 moles of Al and 3 moles of O in 1 mol of the compound.

Let's determine the mass of corundum, by density

Corundum density = Corundum mass / Corundum volume

3.97 g/cm³ = Corundum mass / 15cm³

3.97 g/cm³ . 15cm³ = Corundum mass →59.55 g

Now, let's find out how many moles are in that mass of corundum

1 mol of Al₂O₃ weighs 101.96 g/mol

Then, 101.96 grams of oxide have 2 moles of Al

59.55 g of oxide, will have (59.55 .2)/101.96 = 1.17 moles

Now, we can know the quantity of atoms, by this rule of three

1 mol has NA atoms (6.02×10²³)

1.17 moles will have (1.17 . NA) = 7.03×10²³ atoms of Al

5 0

3 years ago

A lab scientist cools a liquid sample of water (2.6 kg) at 0.00°C to -192°C. The water turns to ice as this temperature change o

soldier1979 [14.2K]

<u>Answer:</u> The heat released for the given process is -1892 kJ

<u>Explanation:</u>

The processes involved in the given problem are:

$1.)H_2O(l)(0^oC,273K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(s)(-192^oC,81K)$

Pressure is taken as constant.

To calculate the amount of heat released at same temperature, we use the equation:

$q=m\times L_{f,v}$ ......(1)

where,

q = amount of heat released = ?

m = mass of water/ice

$L_{f,v}$ = latent heat of fusion or vaporization

To calculate the amount of heat released at different temperature, we use the equation:

$q=m\times C_{p,m}\times (T_{2}-T_{1})$ .......(1)

where,

q = amount of heat released = ?

$C_{p,m}$ = specific heat capacity of medium

m = mass of water/ice

$T_2$ = final temperature

$T_1$ = initial temperature

Calculating the heat absorbed for each process:

<u>For process 1:</u>

Converting the latent heat of fusion in J/kg, we use the conversion factor:

1 kJ = 1000 J

So, $(\frac{-334kJ}{1kg})\times (\frac{1000J}{1kJ})=-334\times 10^3J/kg$

We are given:

$m=2.6kg\\L_f=-334\times 10^3J/kg$

Putting values in equation 1, we get:

$q_1=2.6kg\times (-334\times 10^3J/kg)=-868400J$

<u>For process 2:</u>

We are given:

$m=2.6kg\\C_{p,s}=2050J/kg.K\\T_1=273K\T_2=81K$

Putting values in equation 2, we get:

$q_2=2.6kg\times 2050J/kg.K\times (81-(273))^oC\\\\q_2=1023360J$

Total heat absorbed = $q_1+q_2$

Total heat absorbed = $[-868400+(-1023360)]J=-1891760J=-1891.76kJ\approx -1892kJ$

Hence, the heat released for the given process is -1892 kJ

3 0

3 years ago

You observe 2 waves named alpha and beta. The alpha wave has a frequency of 5 Hz and the beta wave has a frequency of 2 Hz. Base

MAVERICK [17]

Answer:

shorter wavelength = alpha wave

Explanation:

Given that,

The alpha wave has a frequency of 5 Hz and the beta wave has a frequency of 2 Hz.

We need to compare the wavelengths of these two waves.

For alpha wave,

$\lambda_1=\dfrac{c}{f_1}\\\\\lambda_1=\dfrac{3\times 10^8}{5}\\\\=6\times 10^7\ m$

For beta wave,

$\lambda_2=\dfrac{c}{f_2}\\\\\lambda_2=\dfrac{3\times 10^8}{2}\\\\=15\times 10^7\ m$

From the above calculations, we find that the wavelength of the alpha wave is shorter than the wavelength of the beta wave.

8 0

3 years ago