Can you watch the video and help me fill in the blanks ? Thank you

Answer:  

Molarity=moles of solute/ L of solution

Molality = moles of solute/ kg of solvent

Solute= what is being dissolved

Solvent= what is doing the dissolving

Solution= both together

Explanation:

Example's:

#1.  For number one you use the Molarity formula.  M= moles of solute/ L of solution.

To find moles of Mg(NO3)2 you divide 95g by its molar mass which is 148.33g so 95/148.33=.6405 moles of Mg(NO3)2.  Then plug in what you have.  .38M= .6405 moles Mg(NO3)2 / X.  Then solve for X using algebra.  .6405/.38= 1.686 L of solution.  (Volume).

Final Answer: 1.686 L

#2.  For number 2 you use the Molality formula.  m= moles of solute/ kg of solvent.

First you have to find moles of glucose by taking 267g and dividing it by its molar mass which is 180.56g.               267g/180.56g= 1.532 moles of glucose.  Then you have to change L to kg.  The easiest way to do this is to look at the density and see that for every 1 ml there is 1 gram.  So to take Liters to ml you multiply 1.59 by 1000 and get 1590 ml.  So that means you have 1590 grams. then you divide 1590grams by 1000 to get 1.59 Kg of slovent.  Then plug in your information into the formula.      molality= 1.532 moles of glucose / 1.59 Kg of solvent= .964 molality.

Final answer: .964 mol/Kg

#3.  m= moles of solute / Kg of solvent.  0.445 mol solute / 2.07 Kg solvent= .215 Molality

Final Answer: .215 mol/Kg

#4. m= moles of solute / Kg of solvent.  take 13.5g and divide it by ethylene glycols molar mass which is 62.068 g.  13.5g / 62.068g= .218 mol.  Then you take 135g of water and divide it by 1000 to get Kg.  135/1000=.135 Kg.  Then plug in your information.  m= .218mol/.135 Kg= 1.615 molality

Final Answer: 1.615 mol/Kg.

Answer:

E  3.0 x 10² mL.

Explanation:

Hello there!

In this case, according to the formula for the calculation of the mass-volume percent:

[tex]\% m/V=\frac{m_{solute}}{V_{solution}}*100\%[/tex]

Whereas it is necessary to know the mass of the solute and the volume of the solution. Thus, given the mass of NaOH as the solute, the volume of the solution would be:

[tex]V_{solution}=\frac{m_{solute}}{\% m/V}*100\%[/tex]

Then, by plugging in we obtain:

[tex]V_{solution}=\frac{75g}{25\%}*100\%\\\\V_{solution}=3.0x10^2mL[/tex]

Thus, the answer is E  3.0 x 10² mL.

Best regards!

Answer:

B. Salt water has too few water molecules and too many salt molecules

Explanation:

Seawater is toxic to humans because your body is unable to get rid of the salt that comes from seawater. Your body normally gets rid of excess salt by having the kidneys produce urine, but it needs freshwater to dilute the salt in your body for the kidneys to work properly.

Answer:

C

Explanation:

Answer:

C the temperature will increase.

Answer:

We can simply refer to the cation in the ionic compound as magnesium. Phosphorus, Pstart text, P, end text, is a group 15 element and therefore forms 3- anions.

Explanation:

Because it is an anion, we add the suffix -ide to its name to get phosphide as the name of the ion.

Answer:

3.15 × 10⁻⁶ mol H₂/L.s

1.05 × 10⁻⁶ mol N₂/L.s

Explanation:

Step 1: Write the balanced equation

2 NH₃ ⇒ 3 H₂ + N₂

Step 2: Calculate the rate of production of H₂

The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s

Step 3: Calculate the rate of production of N₂

The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:

2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s

Step 1:  When we Write the balanced equation

Step 2: Calculate the rate of production of H₂

Step 3: Calculate the rate of production of N₂

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Answer:

Molecular formula => C₅H₅O₅

Explanation:

From the question given above, the following data were obtained:

Empirical formula = CHO

molar mass of compound = 145 g/mol

Molecular formula =?

The molecular formula of the compound can be obtained as follow:

Molecular formula = Empirical formula × n

Molecular formula = [CHO]ₙ

[CHO]ₙ = 145

[12 + 1 + 16]n = 145

29n = 145

Divide both side by 29

n = 145 / 29

n = 5

Molecular formula = [CHO]ₙ

Molecular formula = [CHO]₅

Molecular formula => C₅H₅O₅

Answer:

2

Explanation:

zeros are not considered sig figs

Answer:

outside the nucleus

Explanation:

Answer:

Two of them are solids, one is liquid. Two of them are edible, one is not. One is a mixture, and two are not.

Explanation:

Answer:

The temperature of the gas is 293.15 K.

The temperature of the gas is 303.15 K.

The new pressure of the gas is 0.97 atm.

Explanation:

You need to add 273.15 to the Celsius temperature to convert from Celsius to Kelvin.

So,

20.0°C + 273.15 = 293.15 K.

30.0°C + 273.15 = 303.15 K.

We can use the following equation to calculate the new pressure:

[tex]P_2 = P_1 *\frac{T_2}{T_1}[/tex]

where:

Plugging in the values from the problem, we get:

[tex]P_2 = 1.00 atm * \frac{303.15 K }{293.15 K} = 0.97 atm[/tex]

Therefore, the new pressure of the gas is 0.97 atm.

Answer: C

waxes and sterols

Explanation:

edge 2021

Answer:

B the weight of the passengers

The answer to your question is true

The statement "The scientific method is a rigid process" is false.

The method by which scientist search for solutions and answers to their problem and question with the help of procedure is called scientific method.

To learn more about scientific method here.

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#SPJ3

Answer:

I think its true I dont really know

Explanation:

true

Answer:

B. Conceiving a child who is a bone marrow match to a living child.

Explanation:

I don't know it just is.

Answer:

N2O5

Explanation:

1. Convert to moles

24.5g N * 1mol/14g = 1.75

70.0g * 1mol/16g = 4.375

2. Divide each value by the smallest

1.75/1.75 = 1

4.375/1.75 = 2.5

3. Multiply each by a whole number so that they are both whole numbers

1*2 = 2

2.5*2 = 5

4. These are moles of elements present in the compound

Answer: N2O5

Question: You decompose a compound containing nitrogen and oxygen in the laboratory and produce 24.5 g of nitrogen and 70.0 g of oxygen. Calculate the empirical formula of the compound.

Answer: true

Explanation:

Answer:

The three main kinds of intermolecular interactions are dipole-dipole interactions, London dispersion forces, and hydrogen bonds

Explanation:

Answer:

[tex]\boxed {\boxed {\sf 2.76 \ mol}}[/tex]

Explanation:

Molarity is found by dividing the moles of solute by liters of solution.

[tex]molarity = \frac {moles}{liters}[/tex]

We know the molarity is 1.2 M (mol\liter) and there are 2.3 liters of solution. Substitute the known values into the formula.

[tex]1.2 \ mol/liter= \frac {x}{2.3 \ liters}[/tex]

Since we are solving for x, we must isolate the variable. It is being divided by 2.3 and the inverse of division is multiplication. Multiply both sides by 2.3 liters.

[tex]2.3 \ liters *1.2 \ mol/liter= \frac {x}{2.3 \ liters}* 2.3 \ liters\\2.3 *1.2 \ mol= x\\2.76 \ mol =x[/tex]

In a solution with a molarity of 1.2 and 2.3 liters of solution, there are 2.76 moles.

Answer:

D is the answer i think

Explanation:

Answer:

34 mL

Explanation:

We'll begin by calculating the number of mole in 1.25 g of sodium carbonate, Na₂CO₃. This can be obtained as follow:

Mass of Na₂CO₃ = 1.25 g

Molar mass of Na₂CO₃ = (23×2) + 12 + (16×3)

= 46 + 12 + 48

= 106 g/mol

Mole of Na₂CO₃ =?

Mole = mass /molar mass

Mole of Na₂CO₃ = 1.25 / 106

Mole of Na₂CO₃ = 0.012 mole

Next, we shall determine the number of mole HCl needed to react with 0.012 mole of Na₂CO₃.

The equation for the reaction is given below:

Na₂CO₃ + 2HCl —> H₂CO₃ + 2NaCl

From the balanced equation above,

1 mole of Na₂CO₃ reacted with 2 moles of HCl.

Therefore, 0.012 mole of Na₂CO₃ will react with = 0.012 × 2 = 0.024 mole of HCl.

Next, we shall determine the volume of HCl required for the reaction. This is illustrated:

Mole of HCl = 0.024 mole

Molarity of HCl = 0.715 M

Volume of HCl =?

Molarity = mole /Volume

0.715 = 0.024 / volume of HCl

Cross multiply

0.715 × volume of HCl = 0.024

Divide both side by 0.715

Volume of HCl = 0.024 / 0.715

Volume of HCl = 0.034 L

Finally, we shall convert 0.034 L to mL

This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.034 L = 0.034 L × 1000 mL / 1 L

0.034 L = 34 mL

Therefore, 34 mL of HCl is needed for the reaction.

The amount of HCl required for counterbalancing 1.25 g of Na2CO3(Sodium Carbonate) would be:

- [tex]34 ml[/tex]

Given that,

Mass of Na2CO3 [tex]= 1.25 g[/tex]

To find the Moles of Na2CO3, we will find the molar mass of Na2CO3,

Molar Mass of or Na2CO3 [tex]= 106 g/mol[/tex]

So,

Moles of Na2CO3 [tex]= mass /molar mass[/tex]

[tex]= 1.25/106[/tex]

[tex]= 0.012 mol[/tex]

To determine the quantity of HCl required to display the reaction with 0.012 mol of Na2CO3

[tex]Na_{2} CO_{2} + 2HCl[/tex] → [tex]H_{2}CO_{3} + 2NaCl[/tex]

While balancing the equation, we know that [tex]0.012[/tex] × [tex]2 = 0.024 mole of HCl[/tex] is necessary to process the reaction.

Now,

As we know,

HCl moles [tex]= 0.024[/tex]

Molarity of HCl [tex]= 0.715 M[/tex]

∵ Quantity of HCl required = Moles/Molarity

[tex]= 0.024 / 0.715[/tex]

[tex]= 0.034 l[/tex] [tex]or 34ml[/tex]

Thus, 34 ml is the correct answer.

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Explanation:

pdrias darme la traduccion no te entiendo

Answer:

3.7 x 10²¹ O₂ molecules

Explanation:

2KClO₃ => 2KCl + 3O₂

given 0.500g KClO₃ => 0.500g/122.55g/mol = 0.0041 mole KClO₃

0.0041 mole KClO₃ => 3/2(0.0041) mole O₂ = 0.0061 mole O₂

0.0061 mole O₂ = 0.0061 mole O₂ x 6.02 x 10²³ molecules O₂/mole O₂

= 3.7 x 10²¹ molecules O₂

Answer:

B.

Explanation:

The metal forms metallic bonds which are more greatly affected by gravity, increasing the mass.

Answer:

For a): The mass percent of naphthalene in the original sample is 31.00 %.

For b): The mass percent of 3-nitroaniline in the original sample is 32.03 %.

For c): The mass percent of benzoic acid in the original sample is 29.97 %.

For d): The total percent recovered is 93.00 %.

Explanation:

Percentage by mass is defined as the ratio of the mass of a substance to the mass of the solution multiplied by 100. The formula used for this is:  

[tex]\text{Percent by mass}=\frac{\text{Mass of substance}}{\text{Mass of a solution}} \times 100[/tex]               ......(1)

a):

Mass of naphthalene = 0.483 g

Mass of the sample = 1.558 g

Plugging values in equation 1:

[tex]\%\text{ mass of naphthalene}=\frac{0.483 g}{1.558}\times 100\\\\\%\text{ mass of naphthalene}=31.00 \%[/tex]

b):

Mass of 3-nitroaniline = 0.499 g

Mass of the sample = 1.558 g

Plugging values in equation 1:

[tex]\%\text{ mass of 3-nitroaniline}=\frac{0.499 g}{1.558}\times 100\\\\\%\text{ mass of 3-nitroaniline}=32.03 \%[/tex]

c):

Mass of benzoic acid = 0.467 g

Mass of the sample = 1.558 g

Plugging values in equation 1:

[tex]\%\text{ mass of benzoic acid}=\frac{0.467 g}{1.558}\times 100\\\\\%\text{ mass of benozic acid}=29.97 \%[/tex]

d):

Total mass recovered = [0.483 + 0.499 + 0.467] = 1.449 g

Mass of the sample = 1.558 g

Plugging values in equation 1:

[tex]\text{Total percent recovered}=\frac{1.449 g}{1.558}\times 100\\\\\text{Total percent recovered}=93.00\%[/tex]