What is the Factor Label Method?

A sailor on a trans-Pacific solo voyage notices one day that if he puts 375. mL of fresh water into a plastic cup fresh water we

ElenaW [278]

Answer:

Amount of salt dissolved in each liter of seawater = 40 g

Explanation:

According to Archimedes principle, a body will float in a fluid if the upthrust experienced by a body is equal to the to the weight of the body.

Also, the volume of seawater displaced equals the volume of freshwater in the cup.

From the above principle, since the freshwater and cup floats in the seawater, their combined weight equals the upthrust.

Therefore, mass of equal volume of displaced seawater = mass of freshwater + mass of cup

Mass of freshwater = density of freshwater * volume

density of freshwater = 1 g/mL; volume = 375 mL

mass of freshwater = 375 mL * 1 g/mL = 375 g

mass of seawater = 375 + 15 = 390 g

mass of salt in 375 mL seawater = mass of seawater - mass of freshwater

mass of salt = (390 - 375) g = 15 g

Since 15 g of salt are dissolved in 375 mL seawater, mass of salt in 1 L of seawater =(1000 mL/ 375) * 15g = 40 g

Therefore, amount of salt dissolved in each liter of seawater = 40 g

3 0

3 years ago

How many neutrons are there in 9 molecules of P4 with mass number 31 and proton number 15?

aalyn [17]

A = mass number which is equal to the total number of protons + number of neutrons, Z= atomic number of any element which is also indicates Number of protons or number of electron in the given atom.

The formula to calculate neutron would be :

mass number - proton

31 - 15 = 16

Thus 16 neutrons are found in one atom of P

There are 4 atom in one molecule

Thus

4 atom /1 molecule * 9 molecule * 16 neutron / 1 atom

= 4*9*16= 576 neutrons

7 0

3 years ago

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Consider the reaction between iodine gas and chlorine gas to form iodine monochloride. A reaction mixture at 298.15k initially c

uranmaximum [27]

Step 1 : Write balanced chemical equation

The balanced chemical equation for the reaction between iodine gas and chlorine gas is given below.

$I_{2} (g) + Cl_{2} (g) -----> 2 ICl (g)$

Step 2 : Set up ICE table

We will set up an ICE table for the above reaction

Following points are considered while drawing ICE table

- The initial concentration of product is assumed as 0

- The change in concentration (C) is assumed as x. Change (x) is negative for reactants and positive for products

- The coefficients in balanced equation are considered while writing C values

Check attached file for ICE table

Step 3 : Set up equilibrium constant equation

The equation for equilibrium constant can be written as

$K_{eq} = \frac{[ICl]^{2}}{[I_{2}][Cl_{2}]}$

Step 4 : Solving for x

Keq at 298.15 K is given as 81.9

Let us plug in the equilibrium values (E) for I₂, Cl₂ and ICl from ICE table

81.9 = $\frac{(2x)^{2}}{(0.437-x) (0.269-x)}$

81.9 = $\frac{(2x)^{2}}{x^{2} -0.706x + 0.118}$

$(2x)^{2} = 81.9 [ x^{2} -0.706x + 0.118]$

$4x^{2} = 81.9x^{2} -57.8x +9.66$

$77.9x^{2} -57.8x+9.66 = 0$

Solving the above equation using quadratic formula we get

x = 0.488 or x = 0.254

The value 0.488 cannot be used because the change (C) cannot be greater that initial concentration of the reactants.

Therefore the change in concentration of the gases during the reaction is 0.254 M

Hence, x = 0.254 M

From the ICE table, we know that the equilibrium concentration of ICl is 2x

$[ICl]_{eq} = 2 ( 0.254) = 0.508 M$

The concentration of ICl when the reaction reaches equilibrium is 0.508 M

6 0

3 years ago

The ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. in t

Debora [2.8K]

The complete balanced chemical equation is:
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g)

In statement form: 4mol NH3 reacts with 5 mol O2 to produce 6 mol H2O

First let us find for the limiting reactant:
>molar mass NH3 = 17 g/mol
moles NH3 = 54/17 = 3.18 mol NH3
This will react with 3.18*5/4 = 3.97 mol O2

>molar mass O2 = 32g/mol
moles O2 = 54/32 = 1.69 mol O2
We have insufficient O2 therefore this is the limiting reactant

From the balanced equation:
For every 5.0 mol O2, we get 6.0 mol H2O, therefore

moles H2O formed = 1.69 mol O2 * 6/5 = 2.025 mol
Molar mass H2O = 18g/mol
<span>mass H2O formed = 2.025*18 = 36.45 grams H2O produced</span>

8 0

3 years ago

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In the compound H2O, the electrons in the bonds are unequally shared between oxygen and hydrogen, forming ____. (1 point)

Aloiza [94]

<span>polar bonds due to high difference . In electronegativity of oxygen and hydrogen.</span>

5 0

3 years ago

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