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valkas [14]
3 years ago
10

Sedimentary rocks are described as small pieces of rock and animal remains that are weathered, swept downstream and then settle,

forming a layer.
A. True
B. False
Chemistry
1 answer:
ss7ja [257]3 years ago
5 0

True,  when small pieces of rock and animal remains weather over the years it sometimes get carried down stream then form a layer of rock. over years different types of rocks settle in the same place and compacts the rocks beneath it forming another layer.

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Write the beta decay equation for the following isotope: 91 38 Sr? Please write out steps
babymother [125]

Answer:

\rm _{38}^{91}Sr \longrightarrow \,  _{-1}^{0}e + \,  _{39}^{91}Y

Explanation:

The unbalanced nuclear equation is

\rm _{38}^{91}Sr \longrightarrow \,  _{-1}^{0}e \, + \, ?

Let's write the question mark as a nuclear symbol.

\rm _{38}^{91}Se} \longrightarrow \,  _{-1}^{0}e \, + \,  _{Z}^{A}X

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.  

Then

98 =  0 + A, so A =  98 -  0 = 98, and

38 = -1 + Z, so Z  = 38 + 1 = 39

Then, your nuclear equation becomes

\rm _{38}^{91}Sr \longrightarrow \,  _{-1}^{0}e + \,  _{39}^{91}X

Element 39 is yttrium, so the balanced nuclear equation is

\rm _{38}^{91}Sr \longrightarrow \,  _{-1}^{0}e + \,  _{39}^{91}Y

7 0
3 years ago
True or false energy has no mass and takes up no space ​
ivanzaharov [21]

Answer:

False

Explanation:

Energy takes up space as mass is a form of energy and energy is always constant.

4 0
3 years ago
Gaseous methane (CH₄) reacts with gaseous oxygen gas (O₂) to produce gaseous carbon dioxide (CO₂) and gaseous water (H₂O) If 0.3
AveGali [126]

Answer : The percent yield of CO_2 is, 68.4 %

Solution : Given,

Mass of CH_4 = 0.16 g

Mass of O_2 = 0.84 g

Molar mass of CH_4 = 16 g/mole

Molar mass of O_2 = 32 g/mole

Molar mass of CO_2 = 44 g/mole

First we have to calculate the moles of CH_4 and O_2.

\text{ Moles of }CH_4=\frac{\text{ Mass of }CH_4}{\text{ Molar mass of }CH_4}=\frac{0.16g}{16g/mole}=0.01moles

\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{0.84g}{32g/mole}=0.026moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CH_4+2O_2\rightarrow CO_2+2H_2O

From the balanced reaction we conclude that

As, 2 mole of O_2 react with 1 mole of CH_4

So, 0.026 moles of O_2 react with \frac{0.026}{2}=0.013 moles of CH_4

From this we conclude that, CH_4 is an excess reagent because the given moles are greater than the required moles and O_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of CO_2

From the reaction, we conclude that

As, 2 mole of O_2 react to give 1 mole of CO_2

So, 0.026 moles of O_2 react to give \frac{0.026}{2}=0.013 moles of CO_2

Now we have to calculate the mass of CO_2

\text{ Mass of }CO_2=\text{ Moles of }CO_2\times \text{ Molar mass of }CO_2

\text{ Mass of }CO_2=(0.013moles)\times (44g/mole)=0.572g

Theoretical yield of CO_2 = 0.572 g

Experimental yield of CO_2 = 0.391 g

Now we have to calculate the percent yield of CO_2

\% \text{ yield of }CO_2=\frac{\text{ Experimental yield of }CO_2}{\text{ Theretical yield of }CO_2}\times 100

\% \text{ yield of }CO_2=\frac{0.391g}{0.572g}\times 100=68.4\%

Therefore, the percent yield of CO_2 is, 68.4 %

6 0
3 years ago
Meiosis I is
ki77a [65]

Answer:

what? am srry hahahahahaha

6 0
3 years ago
starting with 500.4 of glucose, what is the maximum amount of ethanol in grams and i'm liters that can be obtained by this proce
ddd [48]
Hope this helps. Thanks

4 0
3 years ago
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