How many significant figures does 7,874.9640 s have?
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Answer:

40 g= 6.022×10²³

x=1.4×10¹⁵

x=40g×6.022×10²³/1.4×10¹⁵

x=17.77×10⁸

Answer:

V = 11 L

Explanation:

To find the missing value, you need to use the Ideal Gas Law. The equation looks like this:

PV = nRT

In this formula,

 ----> P = pressure (atm)

 ----> V = volume (L)

 ----> n = number of moles

 ----> R = constant (0.0821 atm*L/mol*K)

 ----> T = temperature (K)

You have been given all of the values in the correct units. Therefore, you can plug these values into the equation and simplify to find volume. The answer should have 2 sig figs to match the given value with the lowest amount of sig figs.

PV = nRT

(9.4 atm)V = (4.5 moles)(0.0821 atm*L/mol*K)(286 K)

(9.4 atm)V = 105.6627

V = 11 L

Answer:

I am a small palm-sized pointing device

Answer:

3Fe + 4H₂O --> Fe3O4 + 4H2

Answer:

pH = 4.6  

Explanation:

pH is the negative of the log of the hydrogen ion concentration

- log { 2.52 x 10^-5)   = ~ 4.6

Answer:

In order to determine the charge of these elements you'll have to look at their placement in the periodic table by group number.

Explanation:

To further explain this, there is 18 groups of the periodic table.  These groups include:

Group 1: The alkali metals have a 1+ charge for their entire group.

For example: Sodium would have a charge of 1+.

Group 2: The earth alkali metals all have a charge of 2+. For example Mg has a charge of 2+.

Group 3-12: The transitions metals are a little harder to figure the charge of each, if given the compound they are bonded with you can use that to figure it out but only a few have distinct charges. Such as Ag (Silver) only have a 1+ charge. If you are still trying to figure the charge, try making an electron configuration to figure out it's spare electron count or charge.

Group 13: These elements in group 13 have a typical charge of 3+. For example Al has a charge of 3+.

Group 14: These elements in group 14 have a typical charge of 4+. For example Carbon has a charge 4+

Group 15: These elements start the shift towards a negative charge in the table, so for group 15 the elements have a 3- charge. For example Nitrogen has a 3- charge.

Group 16: These elements continue with a negative in group 16 with a negative charge of 2-. For example Oxygen's charge is 2-.

Group 17: The halogen elements in group 17 have a charge of 1-. For example Chlorine has a charge of -.

Group 18: The noble gases in group 18 do not have a charge because they are some of the most stable elements so they're not looking to gain or lose an electron (That is what a charge is basicaly). So, they're charge is none.  

Answer:

nitrogen compounds in the air

Explanation:

Answer:

1.57

Explanation:

moles = mass / molar mass

moles = 69 / 44.013

moles = 1.57

moles N₂O = 69 : 44 g/mol = 1.568

N₂O : 2 moles N, 1 mole O

moles N = 1.568 x 2 = 3.136

From the calculations, the concentration of the acid is 0.24 M.

The term neutralization has to do with a reaction in which an acid and a base react to form salt and water only.

We have to use the formula;

CAVA/CBVB = NA/NB

CAVANB =CBVBNA

The equation of the reaction is; 2NaOH + H2SO4 ----> Na2SO4 + 2H2O

CA = ?

CB = 1.2 M

VA =  50 mL

VB = 20 mL

NA = 1

NB = 2

CA = CBVBNA/VANB

CA = 1.2 M * 20 mL * 1/ 50 mL * 2

CA = 0.24 M

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7.66g of ethyl butyrate is produced.

The reaction is

CH₃CH₂CH₂COOH + CH₃CH₂OH ----> CH₃CH₂CH₂COOCH₂CH₃

From 0.0845 mol of butanoic acid we get 0.78 x 0.0845 = 0.66 mol of ethyl butyrate.

The molar mass of ethyl butyrate is 116.16g/mol

So 0.66 x 116.16 = 7.66g

7.66g of ethyl butyrate is produced.

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NO is the limiting reagent and 4.34 g is the amount of the excess reagent that remains after the reaction is complete

The reactant that is entirely used up in a reaction is called as limiting reagent.

The reaction:

[tex]2NO(g) +2H_2(g)[/tex] → [tex]N_2 +2H_2O[/tex]

Moles of nitrogen monoxide

Molecular weight: [tex]M_(_N_O_)[/tex]=30g/mol

[tex]n_(_N_O_) =\frac{mass}{molar \;mass}[/tex]

[tex]n_(_N_O_) =\frac{22.0}{30g/mol}[/tex]

[tex]n_(_N_O_) = 0.73 mol[/tex]

Moles of hydrogen

Molecular weight: [tex]M_(_H_2_)[/tex]=30g/mol

[tex]n_(_H_2_) =\frac{mass}{molar \;mass}[/tex]

[tex]n_(_H_2_) =\frac{5.80g}{2g/mol}[/tex]

[tex]n_(_H_2_) = 2.9 mol[/tex]

Hydrogen gas is in excess.

NO is the limiting reagent.

The amount of the excess reagent remains after the reaction is complete.

[tex]n_(_N_2_) =[/tex] (2.9 mol- 0.73 mol NO x [tex]\frac{1 \;mol \;of \;H_2}{2 \;mole \;of \;NO}[/tex]) x [tex]\frac{2g \;of \;H_2}{mole \;of \;H_2}[/tex]

[tex]n_(_N_2_) =[/tex]4.34 g

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Answer:

69.6 g CO₂

Explanation:

To find the mass of carbon dioxide, you need to (1) convert grams LiOH to moles LiOH (via molar mass from periodic table values), then (2) convert moles LiOH to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams CO₂ (via molar mass). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units (desired unit in the numerator). The final answer should have 3 sig figs to reflect the given value's sig figs.

Molar Mass (LiOH): 6.9410 g/mol + 15.998 g/mol + 1.008 g/mol

Molar Mass (LiOH): 23.947 g/mol

2 LiOH(s) + 1 CO₂(g) ----> Li₂CO₃(s) + H₂O(l)

Molar Mass (CO₂): 12.011 g/mol + 2(15.998 g/mol)

Molar Mass (CO₂): 44.007 g/mol

75.8 g LiOH          1 mole             1 mole CO₂               44.007 g
------------------  x  ----------------  x  -----------------------  x  -------------------  =
                            23.947 g         2 moles LiOH              1 mole

= 69.6 g CO₂

The mass in grams of carbon dioxide that would be required to react with 75.8 g of LiOH is 69.3g

The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter.

A mole is defined as the amount of substance containing the same number of atoms, molecules, ions, etc. as the number of atoms in a sample of pure 12C weighing exactly 12 g.

Given,

Mass of LiOH = 75.8g

Moles of LiOH = mass / molar mass

                       = 75.8 / 24 = 3.15 moles

From the reaction,

2 moles of LiOH react with 1 mole of CO₂

1 mole of LiOH needs 0.5 moles of CO₂

thus, 3.15 moles will need 0.5 × 3.15 moles = 1.575 moles of CO₂

Mass of Carbon dioxide = 1.575 × 44

                                       = 69.3g of carbon dioxide

Therefore, the mass in grams of carbon dioxide that would be required to react with 75.8 g of LiOH is 69.3g

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Answer:

Topographic maps generally shows Natural features such as Rivers and Lakes.

sometimes they also show some features made by humans like brodges, elevations etc.

Answer:

Topographic maps show contours, elevation, forest cover, marsh, pipelines, power transmission lines, buildings, and various types of boundary lines such as international, provincial, and administrative, and many others.

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The type of crude oil that is considered to be the highest quality for gasoline is West Texas Intermediate (Option B).

Oil is a fossil fuel obtained from Earth's crust, which is fundamental nowadays to generate electricity.

West Texas Intermediate is a highly pure and high-type energy source of crude oil used in industry.

In conclusion, the type of crude oil that is considered to be the highest quality for gasoline is West Texas Intermediate.

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So mass of sulphur

[tex]\\ \rm\Rrightarrow 0.02\times 757\times 10^{6}[/tex]

[tex]\\ \rm\Rrightarrow 1514\times 10^4[/tex]

Moles =64/32=2

So

So tons of sulphur dioxide

[tex]\\ \rm\Rrightarrow 2(1516)10^4[/tex]

[tex]\\ \rm\Rrightarrow 3032\times 10^4ton[/tex]

Explanation:

2% of 2million tons of sulphur=

2÷100×7.57E9

That is;

[tex] \frac{2}{100} \times 7.57 \times 10 {}^{6} = 0.02 \times 7.57 \times 10 {}^{6} \\ = 2 \times 10 {}^{ - 2} \times7.57 \times 10 {}^{6} = 2 \times 7.57(10 {}^{ - 2 + 6}) \\ = 15.14 \times 10 {}^{4} (for \: sulphur) \\ for \: sulphur \: dioxide = so2 \\ the \: molar \: mass \: of \: so2 = 64g.mol {}^{ -1} \\ if \: 32grams \: of \: sulphur \: weighs \: 15.14 \times 10 {}^{4} tones \\ 64grams \: will \: weigh \: \frac{64}{32} \times 15.14 \times 10 {}^{4} \\ = 2 \times 15.14 \times 10 {}^{4} \\ = 30.28 \times 10 {}^{4} \\ or = 3.028 \times 10 {}^{5} tones[/tex]

Answer:

The answer for this problem would be "b. A 10 mL graduated cylinder".

Explanation:

The reason for this is because with a instrument that is closer to the value you're trying to measure, the more precise it will be.

For example using a teaspoon to measure something rather than a cup.

The atomic mass of the unknown element is; 23

We are given that;

Number of protons = 11

Number of Neutrons = 12

Number of Electrons = 11

Now, formula for atomic mass is;

Atomic mass = Number of Neutrons + Number of Electrons

Atomic Mass = 12 + 11

Atomic Mass = 23

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Carbon dioxide + Water Glucose (sugar) + Oxygen CO2 + H2O C6H12O6 + 6O2 Cellular respiration or aerobic respiration is a series of chemical reactions which begin with the reactants of sugar in the presence of oxygen to produce carbon dioxide and water as waste products.

Answer:

you know that they will be a displacement reaction that will form a barium salt:

Ba(NO3)2+ 2NaCl--> BaCl2 + 2NaNO3

So now that we have that formula and the molecular weight we can determine how much salt will be made. So here we convert the grams to moles

(42.3g Ba(NO3)2)*(1 mole/261.34g) = 0.16185 mol

In the molecular formula we know that 1 mole of Barium nitrate will create 1 mole of Barium chloride, so in this case (in a perfect world) you should get 0.16185 mole of barium chloride (208.23 g/mol) that we then have to convert to grams.

(0.16185 mol BaCl2) * ( 208.23 g/mol) = 33.7037 g of Barium Chloride (rounded to 3 significant digits = 33.7g)

Answer:

4.2  g  

Explanation:

The VOLUME of the ring is   4.2 - 4.0 = .2 ml  = .2 cm^3

the MASS of the ring is this times the density

.2 cm^3  *  21 g/cm^3 = 4.2 g

Answer:

the answer is c

Explanation:

density is mass/volume

so mass=density × volume

but we take the change is volume that is v2-v2=4.2-4=0.2ml

but the density is in gm/cm^3 so we should convert ml into cm^3. eventually they are equal so mass=21×0.2=4.2

The Correct option is A. octahedral

because the molecule has 6 sigma bonds that results into sp³d² hybridization.

and octahedral structure is formed

Answer:

Octahedral

Explanation:

You have a atom connected to 6 others and each of the bonds are similar

Answer:

The si unit used to measure weight is kilogram /kg

Answer:

T₂ = 39.4 °K

Explanation:

Because you are only dealing with volume and temperature, you can use Charles' Law to find the missing value. The formula looks like this:

V₁ / T₁ = V₂ / T₂

In this formula, "V₁" and T₁" represent the initial volume and temperature. "V₂" and "T₂" represent the final volume and temperature. You have been given values for all of the variables except for "T₂". Therefore, by plugging these values into the formula, you can simplify to find the answer.

V₁ = 350 mL                     T₁ = 45 °K

V₂ = 400 mL                     T₂ = ?

V₁ / T₁ = V₂ / T₂                                        <----- Charles' Law formula

(350 mL)(45 °K) = (400 mL)T₂                <----- Insert values into variables

15750 = (400 mL)T₂                                <----- Multiply left side

39.4 = T₂                                                 <----- Divide both sides by 400

Gneiss is a sedimentary rock and granite is a metamorphic rock. That is option A.

Rocks are geological hard materials that are made up of various types which include:

Therefore, Gneiss is a sedimentary rock and granite is a metamorphic rock.

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Answer:

C

Granite is formed due to the slow cooling of magma below Earth's surface and gneiss is formed due to metamorphosis.

Explanation:

I took the test!

For the reactants,

For the products,

Here, arsenic (+5 to +3) and carbon (+3 to +4) are the only oxidation numbers changing.

Note that an increase in oxidation number means electrons are lost. Thus oxidation is occurring, and a decrease in oxidation number means electrons are being gained, and thus reduction is occurring.

Also, the compound that contains the element being oxidized is the reducing agent, and the compound that contains the element being reduced is the oxidizing agent.

So, the answers are:

name of the element oxidized: Carbon

name of the element reduced: Arsenic

formula of the oxidizing agent: [tex]\text{H}_{3}\text{AsO}_{4}[/tex]

formula of the reducing agent: [tex]\text{H}_{2}\text{C}_{2}\text{O}_{4}[/tex]

11.48-gram of [tex]KCl0_3[/tex] are needed to produce 6.75 Liters of [tex]O_2[/tex]  gas measured at 1.3 atm pressure and 298 K

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

First, calculate the moles of the gas using the gas law,

PV=nRT, where n is the moles and R is the gas constant. Then divide the given mass by the number of moles to get molar mass.

Given data:

P= 1.3 atm

V= 6.75 Liters

n=?

R= [tex]0.082057338 \;L \;atm \;K^{-1}mol^{-1}[/tex]

T=298 K

Putting value in the given equation:

[tex]\frac{PV}{RT}=n[/tex]

[tex]n= \frac{1.3 \;atm\; X \;6.75 \;L}{0.082057338 \;L \;atm \;K^{-1}mol^{-1} X 298}[/tex]

Moles = 0.3588 moles

Now,

[tex]Moles = \frac{mass}{molar \;mass}[/tex]

[tex]0.3588 moles = \frac{mass}{32}[/tex]

Mass= 11.48 gram

Hence, 11.48-gram of [tex]KCl0_3[/tex] are needed to produce 6.75 Liters of [tex]O_2[/tex] gas measured at 1.3 atm pressure and 298 K

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This is because fall has more weather patterns which are repetitive than summer.

This is defined as the atmospheric condition of a particular place at a given period of time.

The repetitive weather patterns have various effects on the stars which is why they were studied.

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Answer:

15 oxygens

Explanation:

Given the partially balanced reaction: [tex]\text{C}_{\text{5}}\text{H}_{\text{10}}+\text{O}_{\text{2}} \rightarrow \text{5C} \text{O}_{2}}+\text{5H}_\text{2}\text{O}[/tex]

The subscripts (small number to the right of each element symbol) are the number of atoms of the element within each compound/molecule, and the coefficients (numbers in front of each compound) represent the number of that molecule involved in one full reaction (if the equation were balanced).

The product side of the reaction is on the right of the arrow.

To determine the total number of Oxygens on the product side, we need to identify how many Oxygens are in each molecule (the subscript on the Oxygen), and then multiply times the number of that molecule that would be involved (Coefficient of the compound containing Oxygen).  There are multiple compounds on the right side of the equation that contain Oxygen, so we'll need to add together the number of Oxygens each part contributes.

[tex]\text{C}_{\text{5}}\text{H}_{\text{10}}+\text{O}_{\text{2}} \rightarrow \bold{5}\text{C} \bold{O_{2}}+\bold{5}\text{H}_\text{2}\bold{O}[/tex]

[tex]\text{\# reactant-side Oxygens}=\bold{5}\text{C} \bold{O_{2}}+\bold{5}\text{H}_\text{2}\bold{O}\\=5 \text{ CO}_{\text{2}}\text{ molecules} *\frac{\text{2 Oxygens}}{\text{1 CO}_{\text{2}}\text{ molecule}}+5 \text{ H}_{\text{2}}\text{O molecules} *\frac{1\text{ Oxygen}}{\text{1 H}_{\text{2}}\text{O molecule}}\\=10\text{ Oxygens}+5\text{ Oxygens}\\=15\text{ Oxygens}[/tex]