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3 years ago

1. Brandon is flying to the Western United States. His plane manages to cover 700 miles

Chemistry

1 answer:

torisob [31]3 years ago

4 0

Answer:

a. Bradon was flying at the speed rate of 350 miles/hour.

b. His velocity is 350 miles/hour due west.

Explanation:

The speed of an object is the ratio of distance moved to the time taken, while velocity is the ratio of displacement to time taken. The major difference between speed and velocity is that speed is a scalar quantity, while velocity is a vector quantity.

Vector quantities are described by both magnitude and direction, while scalar have only magnitude.

From the question,

distance = 700 miles and time taken = 2 hours

a. speed = $\frac{distance}{time}$

= $\frac{700}{2}$

= 350 miles/hour

∴ Bradon was flying at the speed rate of 350 miles/hour.

b. velocity = $\frac{displacement}{time}$

displacement = 700 miles of west, time taken = 2 hours

velocity = $\frac{700}{2}$

= 350 miles/hour due west

His velocity is 350 miles/hour due west.

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11Alexandr11 [23.1K]

Answer:

3.11%

Explanation:

We express a solution's volume by volume percent concentration, % v/v,

Take the ratio of the isopropyl alcohol (IPA) volume to the total volume of the solution, which is 1400 mL of water+ 45 mL of IPA,

and multiply by 100 to get the percentage: 45/(1400+45) *100 = 0.0311*100 = 3.11%

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3 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

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3 years ago

Consider the following intermediate reactions.

Alja [10]

2.1648 kg of CH4 will generate 119341 KJ of energy.

Explanation:

Write down the values given in the question

CH4(g) +2 O2 → CO2(g) +2 H20 (g)

ΔH1 = - 802 kJ

2 H2O(g)→2 H2O(I)

ΔH2= -88 kJ

The overall chemical reaction is

CH4 (g)+2 O2(g)→CO2(g)+2 H2O (I) ΔH2= -890 kJ

CH4 +2 O2 → CO2 +2 H20

(1mol)+(2mol)→(1mol+2mol)

Methane (CH4) = 16 gm/mol

oxygen (O2) =32 gm/mol

Here 1 mol CH4 ang 2mol of O2 gives 1mol of CO2 and 2 mol of 2 H2O

which generate 882 KJ /mol

Therefore to produce 119341 KJ of energy

119341/882 = 135.3 mol

to produce 119341 KJ of energy, 135.3 mol of CH4 and 270.6 mol of O2 will require

=135.3 *16

=2164.8 gm

=2.1648 kg of CH4

2.1648 kg of CH4 will generate 119341 KJ of energy

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3 years ago

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A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o

yaroslaw [1]

<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

Mass of the unknown metal sample as 58.932 g
Initial temperature of the metal sample as 101°C
Final temperature of metal is 23.68 °C
Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

Therefore; mass of pure water is 45.2 g
Initial temperature of water = 21°C
Final temperature of water is 23.68 °C
Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

= 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

= 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q = m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

= 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

= 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>

We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.

Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

= 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

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3 years ago

A multivitamin tablet contains 40 milligrams of potassium. how many moles of potassium does each tablet contain?

katrin2010 [14]

Conversion of mole to grams
k in mole = 1 mole/ atomic mass
K in mole =1/ 39.0983 g/mole
= 0.255765 g/mole
converting 40 grams of K
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There are 1.0230623 moles of K in 40 K of Potassium

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3 years ago

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