A sample of a food label is shown below.
Example: Cereal Box
Calories per serving - 167
Total fat - 3 grams
Total saturated fat - 1 gram
Cholesterol - 0mg
Sodium - 250mg
Total carbs - 32 grams
Dietary fiber - 4 grams
Sugars - 11 grams
Protein - 3 grams
Servings per container 5

(a) Is the Calories per serving correctly written? Show your calculation.
(b) Calculate the percent from carbohydrates per serving.
(c) Calculate the Calories if you eat one box of the cereal.​

Answer:

The answer is:

[tex]PART \ A: CH_4\\\\PART \ B: CH_30H\\\\PART \ C: H_2CO[/tex]

Explanation:

Parts A:

The vapor pressure is higher in [tex]CH_4[/tex] because it is non-polar, while [tex]CH_3Cl[/tex] is polar. [tex]CH_4[/tex] has a lower molar weight as well.

Part B:

Although hydrogen bonding is found in both commodities, the vapor pressure is higher because of the smaller molar mass of [tex]CH_30H[/tex].

Part C:

[tex]H_2CO[/tex] does not show hydrogen due to the increased vapor pressure

[tex]CH_3OH[/tex] bonding.

Answer:

0.60 moles of atoms of carbon

Explanation:

Step 1: Given data

Step 2: Calculate the number of carbon atoms in the given sample

According to the chemical formula of the compound, the molar ratio of C to O is 2:6, that is, there are 2 moles of atoms of C every 6 moles of atoms of O. The number of moles of atoms of C is:

1.8 mol O × 2 mol C / 6 mol O = 0.60 mol C

Answer:

200.0lg

Explanation:

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Answer:

ecosystem i think

Answer:

90.26%

Explanation:

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

Theoretical yield of AlCl₃ = 1551 g

Actual yield of AlCl₃ = 1400 g

Percentage yield =?

The percentage yield of the reaction can be obtained as follow:

Percentage yield = Actual yield / Theoretical yield × 100

Percentage yield = 1400 / 1551 × 100

Percentage yield = 140000 / 1551

Percentage yield = 90.26%

Thus, the percentage yield of the reaction is 90.26%

Answer:

Q = -535 J.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the lost energy according to the following and generic heat equation:

[tex]Q=mC(T_F-T_i)[/tex]

Thus, since the specific heat of iron is 0.450 in the SI units, we can plug in the mass and temperatures to obtain:

[tex]Q=28.3g*0.450\frac{J}{g\°C} (24.0\°C-66.0\°C)\\\\Q=-535J[/tex]

Regards

Answer:

I would need to see the diagram in order to answer properly

Answer:

7.15g/cm³

Explanation:

To solve this question we must know that a body-centered cubic unit cell contains 2 atoms.

The volume of the unit cell is:

289pm = (289x10⁻¹²m)³ =

2.414x10⁻²⁹m³ * (1cm³ / 1x10⁻⁶m³) = 2.414x10⁻²³cm³

And the mass is -Molar mass Chromium = 51.9961g/mol:

2atoms * (1mol / 6.022x10²³atoms) * (51.9961g / mol) =

1.727x10⁻²²g

The density is:

1.727x10⁻²²g / 2.414x10⁻²³cm³ =

Answer:

meeting ones legal responsibility

Answer:

50 g Sucrose

Explanation:

Step 1: Given data

Step 2: Calculate the mass of sucrose needed to prepare the solution

The concentration of the solution is 2.5%, that is, there are 2.5 g of sucrose (solute) every 100 g of solution. The mass of sucrose needed to prepare 2000 g of solution is:

2000 g Solution × 2.5 g Sucrose/100 g Solution = 50 g Sucrose

Answer:

P = 487 Torr

mN₂ = 0.347 g

mO₂ = 0.174 g

mHe = 0.0333 g

Explanation:

Step 1: Calculate the total pressure of the mixture

The total pressure of the mixture (P) is equal to the sum of the partial pressures of the gases.

P = pN₂ + pO₂ + pHe

P = 231 Torr + 101 Torr + 155 Torr = 487 Torr

Step 2: Calculate the moles of each gas

We have 3 gases in a 1.00 L container at 25.0 °C (298.2 K). We can calculate the number of moles using the ideal gas equation: P × V = n × R × T.

nN₂ = pN₂ × V / R × T

nN₂ = 231 Torr × 1.00 L / (62.4 Torr.L/mol.K) × 298.2 K = 0.0124 mol

nO₂ = pO₂ × V / R × T

nO₂ = 101 Torr × 1.00 L / (62.4 Torr.L/mol.K) × 298.2 K = 0.00543 mol

nHe = pHe × V / R × T

nHe = 155 Torr × 1.00 L / (62.4 Torr.L/mol.K) × 298.2 K = 0.00833 mol

Step 3: Calculate the mass of each gas

The molar mass of N₂ is 28.01 g/mol.

mN₂ = 0.0124 mol × 28.01 g/mol = 0.347 g

The molar mass of O₂ is 32.00 g/mol.

mO₂ = 0.00543 mol × 32.00 g/mol = 0.174 g

The molar mass of He is 4.00 g/mol.

mHe = 0.00833 mol × 4.00 g/mol = 0.0333 g

Aluminium and Nitrogen!

Answer:

Sample A

Explanation:

Molarity of a solution can be calculated using the formula:

Molarity = number of moles (mol) ÷ volume (vol)

For Sample A:

V = 300ml = 300/1000 = 0.3 L

Molarity = 1M

n = number of moles (mol)

1 = n/0.3

n = 0.3moles

For Sample B:

V = 145 mL = 145/1000 = 0.145L

Molarity = 1.5 M

n = number of moles

1.5 = n/0.145

n = 1.5 × 0.145

n = 0.22 moles

Based on the above results (moles), sample A with 0.3 moles contains the larger concentration of sodium chloride.

Answer:

until the next harvest, and seed must be held for the next season's ... successful grain storage is the moisture content of the crop. ... or both. If ambient temperatures are low, then air alone may cool the ... allow some of the drying to take place naturally while the crop ... employed to cool grain that has been placed in storage.

Answer: The new volume at different given temperatures are as follows.

(a) 109.81 mL

(b) 768.65 mL

(c) 18052.38 mL

Explanation:

Given: [tex]V_{1}[/tex] = 571 mL,       [tex]T_{1} = 26^{o}C[/tex]

(a) [tex]T_{2} = 5^{o}C[/tex]

The new volume is calculated as follows.

[tex]\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL[/tex]

(b) [tex]T_{2} = 95^{o}F[/tex]

Convert degree Fahrenheit into degree Cesius as follows.

[tex](1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C[/tex]

The new volume is calculated as follows.

[tex]\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL[/tex]

(c) [tex]T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C[/tex]

The new volume is calculated as follows.

[tex]\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL[/tex]

Answer:

a. 2.5 g

b. 0.025

c. i. 10 times ii 25 g of KI

Explanation:

a. A 2.5% (by mass) solution concentration signifies that there is ______ of solute in every 100 g

Since % by mass = mass of solute, m/mass of solvent,M × 100 %

2.5 % = m/M × 100%

since M = 100 g,

2.5 % = m/100 × 100 %

m/100 = 2.5/100

m = 2.5 g

b. Therefore, when 2.5% is expressed as a ratio of solute mass over solution mass, that mass ratio would be ________

The mass ratio = mass of solute/mass of solvent

= m/M

= 2.5 g/100 g

= 0.025

c  A solution mass of 1 kg is _______ times greater than 100 g, thus one kilogram (1 kg) of a 2.5% KI solution would contain ____________of

i. A solution mass of 1 kg is _______ times greater than 100 g

Since 1 kg = 1000 g, 1 kg/100 g = 1000 g/100 g = 10

So,1 kg is 10 times greater than 100 g.

ii. Thus one kilogram (1 kg) of a 2.5% KI solution would contain ____________of KI

So, if M = 1 kg = 1000 g, we find m

Since % by mass = mass of solute, m/mass of solvent,M × 100 %

2.5 % = m/M × 100%

since M = 1000 g,

2.5 % = m/1000 × 100 %

m/1000 = 2.5/100

m = 2.5/100 × 1000 g

m = 2.5 × 10 g

m = 25 g

Answer:

Ethylbenzoyl chloride | C9H9ClO

Answer:

Explanation:

Given Information :

Mass = 4.0kg

Acceleration  due to gravity = 10 m/s

Weight= ?

[tex]Weight = \frac{mass}{acceleration\: due \:to \:gravity} \\\\W= \frac{4.0}{10} \\\\W= \frac{2}{5} \\\\W=0.4[/tex]

Answer:

Explanation:

The boiling point will increase due to dissolution of sugar in water . Increase in boiling point ΔT

ΔT = Kb x m , where Kb is molal elevation constant water , m is molality of solution

Kb for water = .51°C /m

moles of sugar = 16.90 / 342.3

= .04937 moles

m = moles of sugar /  kg of water

= .04937 / .04090

= 1.207

ΔT = Kb x m

= .51 x 1.207

= .62°C .

So , boiling point of water = 100.62°C .

Unnillium (101)

Unnilbium (102)

Unniltrium (103)

Unnilquadium (104)

Unnilpentium (105)

Unnilhexium (106)

Unnilseptium (107)

Unniloctium (108)

Unnilennium (109)

Ununnillium (110)

Answer:

Sample B

Explanation:

In this case, we need to determine the empirical formula for each sample. The one that match the formula of the propene would be the sample.

Let's do Sample A:

C: 60 g;       H: 12 g

1. Calculate moles:

We need the atomic weights of carbon (12 g/mol) and hydrogen (1 g/mol):

C: 60 / 12 = 5

H: 12 / 1 = 12

2. Determine number of atoms in the formula

In this case, we just divide the lowest moles obtained in the previous part, by all the moles:

C: 5 / 5 = 1

H: 12 / 5 = 2.4    or rounded to two

3. Write the empirical formula:

Now, the prior results, represent the number of atoms in the empirical formula for each element, so, we put them with the symbol and the atoms as subscripted:

C₁H₂ = CH₂

Therefore, sample A is not the same as propene.

Sample B:

C: 72 g    H: 12 g

Following the same steps, let's determine the empirical formula for this sample

C: 72 / 12 = 6 ---> 6 / 6 = 1

H: 12 / 1 = 12 ----> 12 / 6 = 2

EF: CH₂

Sample C:

C: 84 g    H: 10 g

C: 84 / 12 = 7 ----> 7 / 7 = 1

H: 10 / 1 = 10    ----> 10 / 7 = 1.4 or just 1

EF: CH

Sample D

C: 90 g      H: 10 g

C: 90 / 12 = 7.5     -----> 7.5 / 7.5 = 1

H: 10 / 1 = 10  -------> 10 / 7.5 = 1.33 or just 1

EF: CH

Neither compound has the same empirical formula as C3H6, but C3H6 is a molecular formula, so, if we just simplify the formula we have:

C3H6  -----> CH₂

Therefore, sample B is the one that match completely. Sample B would be the one.

Hope this helps

Answer: When exhaust from car engines spreads throughout our atmosphere and adds to pollution then it is an example of increase in entropy.

Explanation:

A degree of randomness in the molecules of a substance is called entropy.

When exhaust from car engines spreads throughout our atmosphere then it means the molecules are moving at a faster speed due to which they spread into the atmosphere rapidly.

This means that the entropy of exhaust is increasing.

Thus, we can conclude that when exhaust from car engines spreads throughout our atmosphere and adds to pollution then it is an example of increase in entropy.

An increase in entropy

Answer:

203 grams

Explanation:

The no. of moles of (6.3 x 10²⁴ molecules--Avagadros number) of NH₃ = (1.0 mol)(7.2 x 10²⁴ molecules)/(6.022 x 10²³ molecules) = 11.96 mol.

The no. of grams of NH₃ present = no. of moles x molar mass = (11.96 mol)(17.0 g/mol) = 203.3 g ≅ 203.0 g.

The answer is the mass of a sample of NH3 containing 7.20 x 1024 molecules of NH3 is 203 grams.

A molecule is a group of two or more atoms that form the smallest identifiable unit into which a pure substance can be divided and still retain the composition and chemical properties of that substance.

6.02214076 × 10²³  molecules constitutes 1 mole of NH₃

7.20 x 10²⁴ molecules constitutes of

[tex]\rm\dfrac{7.20 \times 10^{24}}{6.02\times 10^{23}}\\[/tex]

=11.19 moles

1 mole of NH₃ = 17 gm

11.19 mole will be = 17 * 11.19

= 203 grams

Therefore the mass of a sample of NH3 containing 7.20 x 1024 molecules of NH3 is  203 grams.

To know more about molecule

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Answer:

mass of sucrose = 17.115 grams

Explanation:

Given that:

mass = 2000 grams

molar mass of sucrose = 342.3 g/mol

2.5% of the sucrose soltion.

Let assume we are given 1 M of the sucrose solution;

2.5% of 1 M = 0.025 M

∴

Molarity = ( mass/molar  mass )(1000/V)

mass = (0.025 * 342.3 * 2000)/(1000)

mass of sucrose = 17.115 grams

Answer:

B.Anything that takes up space and has mass

Answer:

What about it? I don't get the question