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Likurg_2 [28]
3 years ago
14

Can someone help me here lol

Chemistry
2 answers:
Romashka-Z-Leto [24]3 years ago
8 0

Answer:

3) option c

4) option b

5) option a

<u><em>pls mark brainliest</em></u>

icang [17]3 years ago
6 0
The person above is correct
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Layers of rock containing fossils, like the layers illustrated here, are MOST LIKELY composed of ______________ rocks. A) igneou
Ilya [14]
The answer should be B:Sedimentry
8 0
3 years ago
The pH of a vinegar solution is 4.15. What is the H+ concentration of the solution
marishachu [46]
<span>The pH of a vinegar solution is 4.15. To find the H+ concentration of the solution use the following equation -log(H+)=pH. Insert the pH into the equation to get, -log(H+) = 4.15 Rearrange the equation to get, 10^(-4.15) = H+ Finally, you can solve for H+. The hydrogen ion concentration of the vinegar solution is .0000708 M.</span>
3 0
3 years ago
Read 2 more answers
What are the empirical and molecular formulas of a hydrocarbon if combustion of 2.10 g of the compound yields 6.59 g co2 and 2.7
mariarad [96]

 The  empirical  formula    of hydrocarbon is  CH2

The  molecular formula  of the  hydrocarbon is  C6H12


    <u><em>Explanation</em></u>

Hydrocarbon  is  made up  of carbon and hydrogen


<h3><u><em> </em></u>Empirical formula  calculation</h3>

 Step 1:  find  the  moles   CO2  and  H2O

moles =mass/molar mass

moles   of CO2 =  6.59 g/ 44 g/mol = 0.15 moles

moles of H2O = 2.70 g / 18 g/mol =  0.15  moles

Step 2: Find the moles  ratio  of Co2:H2O  by diving  each mole by smallest mole(0.15)

that  is  for  CO2 = 0.15/0.15  =1

              For H2O = 0.15/0.15 =1

therefore  the mole ratio  of Co2 : H2O = 1:1  which  implies that 1 mole of Co2  and 1  mole of H2O is  formed  during combustion reaction.


From the  the law of mass conservation the number  of atoms in reactant side  must  be equal to  number of  atoms  in product side

therefore  since  there 1 atom  of C  in product side there  must be 1 atom of C  in reactant  side.

In addition  there is 2 H atom in product  side  which should be the  same  in reactant side.  

From information above the empirical formula is therefore = CH2


Molecular formula  calculation

[CH2}n= 84 g/mol

[12+ (1x2)] n = 84 g/mol

14 n =  84 g/mol

n = 6

multiply the  each subscript  in CH2  by  6

 Therefore the molecular formula = C6H12




5 0
3 years ago
Galena, a mineral of lead, is a compound of the metal with sulfur. Analysis shows that a 2.34-g sample of galena contains 2.03 g
dolphi86 [110]

The mass of sulfur in the sample is 0.31g

To calculate mass of sulfur in sample: given that amount galena is 2.34g and mass of lead is 2.03 g as galena contain both lead and sulfur the mass of sulfur = mass of galena- mass of lead , that is, 2.34- 2.03=0.31g.The molecular mass is a algebraic sum of atomic mass of each element in the compound thus the mass of the compound can be obtained by adding atomic of each element. Lead(II) sulphide occurs naturally as the mineral galena, often known as lead glance (PbS). It is a significant source of silver and the most significant lead ore.

To learn more about mass:

brainly.com/question/19694949

#SPJ4

7 0
2 years ago
What is the natural and human factors that affect climate?
frozen [14]

Answer:

Some of the various things that humans do to the environment is pump CO2 into the air with Fossil Fuels and using Power Plants, Cutting down trees make more CO2 which can make the Earth hotter and cause Global Warming. Some natural things that can tend to affect climate can be changes in Volcanic activity, Solar activity with the sun, and Earth's orbit around the sun.

Explanation:

It may seem crazy how Earth's orbit around the sun make climate change, but how else would 4 seasons exist. The orbit tends to be in an oval shape. The top parts of the oval, would represent winter and fall because of the distance between the sun decreases. The closer sides of the oval however, would represent Summer and spring because those are the warmer months and show that Earth is closer to the sun.

8 0
3 years ago
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