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3 years ago

Helppp.....Which one of the following statements does not describe gold?

Chemistry

1 answer:

Mariulka [41]3 years ago

4 0

Answer: D. It reacts with oxygen in the presence of moist air.

Explanation:

The element Gold, does not react with oxygen,hence it is not affected by corrosion,neither does it rust.

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Three things about the body systems?

makvit [3.9K]

Answer:

Three things about our body's systems:

All systems have a method of self-regulation or exogenous regulation by other systems.

All systems have a balance in their functions.

All the systems of our organism are intertwined with each other thus giving general vitality.

Explanation:

Best known systems:

Renal, respiratory, circulatory, cardiac, nervous, immune, blood, muscular systems.

All of them include the participation of one or more organs

5 0

3 years ago

A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?

Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

$\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles$

Now we have to calculate the moles of $Br_2$

${\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles$

${\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles$

The balanced chemical reaction is,

$2Na(s)+Br_2(g)\rightarrow 2NaBr$

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium bromide

Now we have to calculate the percent yield of reaction

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%$

Therefore, the percent yield is, 93.4%

3 0

3 years ago

In a sample containing 100 atoms of X, 70 were found to be 9X while 30 were 11X isotopes. Determine the average relative atomic

Burka [1]

The average relative atomic mass of a sample containing 100 atoms of X, 70 were found to be 9X while 30 were 11X isotopes is 9.6g.

<h3>What is Atomic mass? </h3>

Atomic mass is defined as the whole mass of an atom.

It is also defined as the sum of atomic number and number of neutrons.

Atomic mass = Atomic number + neutrons

<h3>What is Isotopes?</h3>

Isotopes are the those element which have same atomic number but have different mass number and number of neutrons.

The average relative atomic mass can be calculated as

mass of isotopes/ mass of sample

mass of all isotopes = (70 × 9X) + (30 × 11X)

=(630 + 330) X

= 960X

Average relative atomic mass = 960X/ 100 X

= 9.6 g

Thus, we concluded that the average relative atomic mass of a sample containing 100 atoms of X, 70 were found to be 9X while 30 were 11X isotopes is 9.6g.

learn more about atomic mass:

brainly.com/question/14250653

#SPJ9

5 0

1 year ago

Glucose, C 6 H 12 O 6 , is used as an energy source by the human body. The overall reaction in the body is described by the equa

MA_775_DIABLO [31]

Answer:

$m_{O_2}=61.87gO_2$

$m_{CO_2}=85.07gCO_2$

Explanation:

Hello,

Considering the given reaction's stoichiometry, grams of oxygen result:

$m_{O_2}=58.0gC_6H_{12}O_6*\frac{1molC_6H_{12}O_6}{180gC_6H_{12}O_6}*\frac{6molO_2}{1molC_6H_{12}O_6}*\frac{32gO_2}{1molO_2}\\m_{O_2}=61.87gO_2$

Moreover, the mass of produced carbon dioxide turns out:

$m_{CO_2}=58.0gC_6H_{12}O_6*\frac{1molC_6H_{12}O_6}{180gC_6H_{12}O_6}*\frac{6molCO_2}{1molC_6H_{12}O_6}*\frac{44gCO_2}{1molCO_2}\\m_{O_2}=85.07gCO_2$

Best regards.

6 0

3 years ago

Pls help ASAP. I will give brainliest

Elena-2011 [213]

Answer:

Explanation:

8 0

3 years ago

Read 2 more answers