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trapecia [35]
3 years ago
7

Mr. McPhillip has a ribbon of magnesium (Mg), a silvery metal. As a demonstration for his class, he sets the metal on fire. It b

urns brightly, leaving behind a whitish powder.
Mr. McPhillip has a ribbon of magnesium (Mg), a silvery metal. As a demonstration for his class, he sets the metal on fire. It burns brightly, leaving behind a whitish powder.

The magnesium is a(n) _______, and the powder produced by burning it is a(n) _______.
A.
element; compound
B.
compound; compound
C.
compound; element
D.
element; element
Chemistry
2 answers:
ICE Princess25 [194]3 years ago
8 0
Magnesium is an element and magnesium oxide is a compound, so Answer A.
OLEGan [10]3 years ago
5 0
A. Magnesium is an element and after it is burned it is a compound
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A mixture of gases with a pressure of 800.0 mm hg contains 60% nitrogen and 40% oxygen by volume. what is the partial pressure o
klasskru [66]
Hello!

<span>We have the following statement data:
</span>
Data:
P_{Total} = 800 mmHg
P\% N_{2} = 60\%
P\% O_{2} = 40\%
P_{partial} = ? (mmHg)

<span>As the percentage is the mole fraction multiplied by 100:

</span>P =  X_{ O_{2} }*100

<span>The mole fraction will be the percentage divided by 100, thus:
</span><span>What is the partial pressure of oxygen in this mixture? 
</span>
X_{ O_{2} }  =  \frac{P}{100}
X_{ O_{2}} =  \frac{40}{100}
\boxed{X_{ O_{2}} = 0.4}


<span>To calculate the partial pressure of the oxygen gas, it is enough to use the formula that involves the pressures (total and partial) and the fraction in quantity of matter:
</span>
In relation to O_{2} :

\frac{P O_{2} }{P_{total}} = X_O_{2}
\frac{P O_{2} }{800} = 0.4
P_O_{2} = 0.4*800
\boxed{\boxed{P_O_{2} = 320\:mmHg}}\end{array}}\qquad\quad\checkmark
<span>
Answer:
</span><span>b. 320.0 mm hg </span>
7 0
3 years ago
5. An example of a muscle that is voluntarily controlled is a muscle that
Shtirlitz [24]

Answer:

A, Makes the leg move

Explanation:

6 0
2 years ago
A horizontal pipe 15 cm in diameter and 4 m long is buried in the earth at a depth of 20 cm. The pipe-wall temperature is 75 deg
Basile [38]

Explanation:

The given data for case (1) is as follows.

  h = 20 cm = 0.2 m

Assuming that a rectangular slab is placed above the pipe and we will calculate the heat transfer as follows.

              Q = kA \frac{\Delta T}{L}

  where,    A = area

                  L = length

                  k = thermal conductivity = 0.8 W/m

             \Delta T = change in temperature.

Therefore, putting the given values into the above formula as follows.

            Q = kA \frac{\Delta T}{L}

                = 0.8 W/m \times (4 m \times 0.15 m) \frac{75 - 5}{0.2}

                = 168 W

For case (2), h = 180 cm = 1.8 m

Therefore, heat lost will be calculated as follows.

                       Q = kA \frac{\Delta T}{L}

                           = 0.8 W/m \times (4 m \times 0.15 m) \frac{75 - 5}{1.8}

                           = 18.67 W

Thus, we can conclude that 18.67 W heat lost if the pipe was buried at a depth of 180 cm.

8 0
3 years ago
The theoretical yield of NaBr from
Dvinal [7]

the percent yield of the reaction is 100%.

The percent yield is calculated as the experimental yield divided by the theoretical yield x 100%:

% yield = actual yield / theoretical yield * 100%

% yield of a reaction in this case Rate

In this case, the molar mass of NaBr is 102.9 g / mol, as you know:

444 actual yield = 7.08 mol x 102.9 g / mol = 728.532 g

theoretical yield = 7.08 mol x 102.9 g / mol = 728.532 g

, Replaced by the definition of percent yield:

percent yield = 728.532 grams / 728.532 grams * 100%

percent yield = 100%

Finally, the percent yield of the reaction is 100%.

<h3 />

FeBr3 is iron bromide. Also known as iron bromide. Iron bromide is an ionic compound in which iron is in a +3 oxidation state.

Learn more about % yield here:brainly.com/question/27979178

#SPJ10

5 0
2 years ago
What is the frequency of radiation whose wavelength is 2.25 X 10^-5cm?
Semenov [28]

Answer:

1.25 x 10^15Hz

Explanation:

c = frequency x wavelength

c is the speed of light, which is equal to 3.00 x 10^8 m / s

frequency = c /wavelength

= (3.00 x 10^8m /s) / (2.40 x 10^-5 cm x 1 m /100cm)

= (3.00 x 10^8 m/s) / 2.40 x 10^-7m

= 1.25 x 10^15/s 1 / s = 1Hz

So, the Frequency = 1.25 x 10^15Hz

I hope this helped :)

3 0
3 years ago
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