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3 years ago

Which element of the scientific process includes the greatest amount of information about living things?

1 answer:

7 0

I would say its hydrogen, as it makes 51% of the cell constituents. This is followed by oxygen, which makes up 24% of the cell, and then carbon which is present as 11% of the total mass. Nitrogen makes up 4% of the cell, followed by other elements like phosphorous.

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Which statement describes the use of a flowchart?

castortr0y [4]

Answer: I believe it is B

5 0

3 years ago

Read 2 more answers

Which of the following produces a physical change in the shape of an object?

bogdanovich [222]

Answer:Any change that occurs without altering the chemical composition of a substance is a physical change. Physical changes can include changing the color, shape, state of matter, or volume of a substance. It is crucial to remember that physical changes never alter the chemical makeup.

Explanation:

i hope that helps u try to figure it out a little bit sorry i couldn't find your answer i didn't have much to go off of

8 0

3 years ago

Compute the values. Express these answers to the hundredths place (i.e., two digits after the decimal point). log ( 2.1 ) = ln (

zepelin [54]

Answer:

x = 4.17

y = 1.86

Explanation:

0.62 = log(x)

x = 10^0.62 = 4.17 ( to the nearest hundredth)

0.62 = ln(y)

y = e^0.62 = 1.86 (to the nearest hundredth)

8 0

3 years ago

Molarity to percent by mass. Convert 1.672 mol/L MgCl2(aq) solution to percent by mass of MgCl2 in the solution. The solution de

nignag [31]

Answer:

$\%m/m=14\%$

Explanation:

Hello!

In this case, since the molarity of magnesium chloride (molar mass = 95.211 g/mol) is 1.672 mol/L and we know the density of the solution, we can first compute the concentration in g/L as shown below:

$[MgCl_2]=1.672\frac{molMgCl_2}{L}*\frac{95.211gMgCl_2}{1molMgCl_2}=159.2\frac{gMgCl_2}{L}$

Next, since the density of the solution is 1.137 g/mL, we can compute the concentration in g/g as shown below:

$[MgCl_2]=159.2\frac{gMgCl_2}{L}*\frac{1L}{1000mL}*\frac{1mL}{1.137g}=0.14$

Which is also the by-mass fraction and in percent it turns out:

$\%m/m=0.14*100\%\\\\\%m/m=14\%$

Best regards!

6 0

3 years ago

Consider the reaction 2CO * O2 —> 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c

Arturiano [62]

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

$m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO} *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2$

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

$Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%$

Best regards!

8 0

3 years ago