Please help!!!

Which of the following is true?

Umnica [9.8K]

Answer:

The answer should be D

Explanation:

because turning 1 mole of propane to grams means its still one mole of propane just in a different unit.

8 0

2 years ago

A sample of food containing 27 g of fat, 48 g of carbohydrates and 20 g of protein is burned in a bomb calorimeter. In a perfect

rosijanka [135]

Answer:

38.3958 °C

Explanation:

As,

1 gram of carbohydrates on burning gives 4 kilocalories of energy

1 gram of protein on burning gives 4 kilocalories of energy

1 gram of fat on burning gives 9 kilocalories of energy

Thus,

27 g of fat on burning gives 9*27 = 243 kilocalories of energy

20 g of protein on burning gives 4*20 = 80 kilocalories of energy

48 gram of carbohydrates on burning gives 4*48 = 192 kilocalories of energy

Total energy = 515 kilocalories

Using,

$Q=m_{water}\times C_{water}\times (T_f-T_i)$

Given: Volume of water = 23 L = 23×10⁻³ m³

$Density=\frac{Mass}{Volume}$

Density of water= 1000 kg/m³

So, mass of the water:

$Mass\ of\ water=Density \times {Volume\ of\ water}$

$Mass\ of\ water=1000 kg/m^3 \times {0.023\ m^3}$

Mass of water = 23 kg

Initial temperature = 16°C

Specific heat of water = 0.9998 kcal/kg°C

$515=23\times 0.9998\times (T_f-16)$

Solving for final temperature as:

<u>Final temperature = 38.3958 °C </u>

8 0

3 years ago

Which ions aren't shown in net ionic reactions, because they are present as both reactants and products of the reaction, so they

Anvisha [2.4K]

Explanation:

This is correct!

Ions that exist in both the reactant and product side of the equation are referred to as spectator ions. Overall, they do not partake in the reaction. If they are present on both sides of the equation, you can cancel them out.

An example is;

Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq) → Na+(aq) + NO3−(aq) + AgCl(s)

The ions; Na+, NO3−(aq) would be cancelled out to give;

Cl−(aq) + Ag+(aq) → AgCl(s)

7 0

2 years ago

Dosage calculation order: 3 mg available: 2 mg per 6 ml how many ml will be given?

Anna35 [415]

9ml will be given for the case of dosage calculation order: 3 mg available: 2 mg per 6 ml

Conversion factors are necessary for dosage calculation, such as when translating from pounds to kilograms or liters to milliliters. This approach, which is straightforward in design, enables physicians to deal with different units of measurement and convert factors to arrive at the solution.

dosage calculation techniques serve as a second or third check on the accuracy of the previous computation techniques. Dimensional Analysis, Ratio Proportion, and Formula or Desired Over Have Method are the three main approaches for dosage calculation. dosage calculations are frequently prescribed and labeled based on their weight or, for solutions, their strength, which is the amount of weight dissolved or suspended in a given volume.

To learn more about dosage calculation please visit -
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3 0

1 year ago

What will the concentration of PCl5 be when equilibrium is reestablished after addition of 1.31 g Cl2? PCl5(g) ⇆ PCl3(g) + Cl2(g

Masteriza [31]

Answer:

The new concentration of PCl5 will be 0.01953 M

Explanation:

Step 1: Data given

Mass of Cl2 added = 1.31 grams

Molar mass Cl2 = 70.9 g/mol

Original Equilibrium Mixture:

3.42 g PCl5

4.86 g PCl3

3.59 g Cl2

Volume = 1.0 L

Step 2: The balanced equation

PCl5(g) ⇆ PCl3(g) + Cl2(g)

Step 3: Calculate the original moles and molarity

Moles = mass / molar mass

Moles PCL5 = 3.42 grams / 208.24 g/mol

Moles PCl5 = 0.0164 moles

[PCl5] = 0.0164 M

moles PCl3 = 4.86 grams / 137.33 g/mol

moles PCl3 = 0.0354 moles

[PCl3] = 0.0354 M

moles Cl2 = 3.59 grams / 70.9 g/mol

moles Cl2 = 0.0506 moles

[Cl2] = 0.0506 M

the new mass Cl2 = 3.59 + 1.31 = 4.9 grams

moles Cl2 = 0.0691 moles

[Cl2]= 0.0691 M

The new concentration at the equilibrium

[PCl5] = 0.0164 + X M

[PCl3 ] = 0.0354 - X M

[Cl2] = 0.0691 - X M

Step 4: Calculate Kc

Kc = [Cl2][PCl3] / [PCl5]

Kc = (0.0506*0.0354)/0.0164

Kc = 0.109

Step 5: Calculate [PCl5]

Kc = 0.109 = ((0.0691 - X)(0.0354 - X)) / (0.0164 + X)

X = 0.00313

[PCl5] = 0.0164 + 0.00313 M = 0.01953 M

[PCl3 ] = 0.0354 - 0.00313 M = 0.03227 M

[Cl2] = 0.0691 - 0.00313 M = 0.06597

The new concentration of PCl5 will be 0.01953 M

6 0

3 years ago