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2 years ago

What would you see if you were standing on the Earth's surface in the penumbra and you were looking up at the sky?

2 answers:

5 0

If you are standing in the Moon's penumbra and look at the Sun, you will witness a partial solar eclipse. During some eclipses, the Sun, the Moon, and the Earth do not form a perfectly straight line, so only the penumbra falls on the Earth's surface while the umbra, the shadow's dark center, is cast into space. please mark me brainliest i lost my other account :(

Solnce55 [7]2 years ago

3 0

Answer:

clouds?

Explanation:

im sorry i just wanted the 23 points

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Double-click on a sheet tab to _______ the sheet.

Vika [28.1K]

Copy it I think hp[e that is right lol

7 0

3 years ago

Read 2 more answers

A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm

Ulleksa [173]

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100$

$Percent error = \frac{0.1}{63.2}\times 100$

$Percent error = 0.00158\times 100$

Percent error = 0.158 %

2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100$

$Percent error = \frac{0.2}{63.2}\times 100$

$Percent error = 0.00316\times 100$

Percent error = 0.316 %

3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100$

$Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100$

$Percent error = \frac{(0.5)}{63.2}\times 100$

$Percent error = 0.00791\times 100$

Percent error = 0.791 %

Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%

7 0

3 years ago

An electrochemical cell is composed of pure nickel and pure iron electrodes immersed in solutions of their divalent ions at room

Andrej [43]

Answer:

0.758 V.

Explanation:

Hello!

In this case, case when we include the effect of concentration on an electrochemical cell, we need to consider the Nerst equation at 25 °C:

$E=E\°-\frac{0.0591}{n} log(Q)$

Whereas n stands for the number of moles of transferred electrons and Q the reaction quotient relating the concentration of the oxidized species over the concentration of the reduced species. In such a way, we can write the undergoing half-reactions in the cell, considering the iron's one is reversed because it has the most positive standard potential so it tends to reduction:

$Fe^{2+}+2e^-\rightarrow Fe^0\ \ \ E\°=0.440V\\\\Ni^0\rightarrow Ni^{2+}+2e^-\ \ \ E\°=-0.250V$

It means that the concentration of the oxidized species is 0.002 M (that of nickel), that of the reduced species is 0.40 M and there are two moles of transferred electrons; therefore, the generated potential turns out:

$E=(0.440V+0.250V)-\frac{0.0591}{2} log(\frac{0.002M}{0.40M} )\\\\E=0.758V$

Beat regards!

8 0

3 years ago

Consider the reaction of NO and CO to form N2 and CO2, according to the balanced equation: 2 NO (g) + 2 CO (g) → N2 (g) + 2 CO2

Gekata [30.6K]

The image is not given in the question, it is attached below:

<u>Answer:</u> The excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.

<u>Explanation:</u>

In the given image:

Red spheres represent oxygen atoms, blue spheres represent nitrogen atoms and black spheres represent carbon atoms

The combination of 1 black and 2 red spheres will represent carbon dioxide $(CO_2)$ compound

The combination of 2 blue spheres will represent nitrogen molecule $(N_2)$

The combination of 1 blue and 1 red sphere will represent nitrogen monoxide $(NO)$ compound

The combination of 1 black and 1 red sphere will represent nitrogen monoxide $(NO)$ compound

Limiting reagent is defined as the reagent which is completely consumed in the reaction and limits the formation of the product.

Excess reagent is defined as the reagent which is left behind after the completion of the reaction.

We are given:

Given moles of NO = 6 moles

Given moles of CO = 4 moles

For the given chemical equation:

$2NO(g)+2CO(g)\rightarrow N_2(g)+2CO_2(g)$

By stoichiometry of the reaction:

If 2 moles of CO reacts with 2 moles of NO

So, 4 moles of CO will react with = $\frac{2}{2}\times 4=4mol$ of NO

As the given amount of NO is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, CO is considered a limiting reagent because it limits the formation of the product.

Hence, the excess reactant is NO, the limiting reactant is CO and the products are shown in the image attached below.

3 0

3 years ago

The asthenosphere is the part of the upper mantle just below the lithosphere that os involved in the plate tectonic movement anf

likoan [24]

Answer:C

Explanation:

3 0

3 years ago