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Montano1993 [528]
4 years ago
7

How many atoms are in 187 g of Calcium?

Chemistry
1 answer:
N76 [4]4 years ago
4 0
Atomic mass Calcium ( Ca) = 40.078 u.m.a

40.078 g --------------- 6.02x10²³ atoms
187 g ------------------- ??

187 x ( 6.02x10²³) / 40.078 =

1.125x10²⁶ / 40.078 = 2.808x10²⁴ atoms

hope this helps!




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2. A quantity of 1.922g of methanol (CH3OH) was burned in a constant-volume
Cerrena [4.2K]
Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
5 0
3 years ago
Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0
3 years ago
What type of rock is show here? How do you know?
mart [117]

Answer:

Sedimentary Rock. You know from the fossils and remains of shells inside.

Explanation:

If it were igneous, the fossils would be melted away. Maybe not with metamorphic, but it's most likely sedimentary.

5 0
4 years ago
Analysis of an athletes urine found the presence of a compound with a molar mass of 312 g/mol. How many moles of this compound a
rewona [7]
<h3>Answer:</h3>

= 5.79 × 10^19 molecules

<h3>Explanation:</h3>

The molar mass of the compound is 312 g/mol

Mass of the compound is 30.0 mg equivalent to 0.030 g (1 g = 1000 mg)

We are required to calculate the number of molecules present

We will use the following steps;

<h3>Step 1: Calculate the number of moles of the compound </h3>

Moles=\frac{mass}{molar mass}

Therefore;

Moles of the compound will be;

=\frac{0.030}{312g/mol}

      = 9.615 × 10⁻5 mole

<h3>Step 2: Calculate the number of molecules present </h3>

Using the Avogadro's constant, 6.022 × 10^23

1 mole of a compound contains 6.022 × 10^23  molecules

Therefore;

9.615 × 10⁻5 moles of the compound will have ;

= 9.615 × 10⁻5 moles × 6.022 × 10^23  molecules

= 5.79 × 10^19 molecules

Therefore the compound contains 5.79 × 10^19 molecules

5 0
3 years ago
I neeeed helpp please
Ymorist [56]
4= lemon juice contains CITRIC ACID and represents a pH of 2.3
6 0
3 years ago
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