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2 years ago

Vibrations that move through the ground carrying the energy released during an earthquake are called

2 answers:

5 0

Vibrations that move through the ground carrying the energy released during an earthquake are called seismic waves. Seismic waves can be measured using a seismometer, which record the motion of the ground.

QveST [7]2 years ago

5 0

They are Called Siesmic Waves. It is Number 3

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A 0.1326 g sample of magnesium was burned in an oxygen bomb calorimeter. the total heat capacity of the calorimeter plus water w

Sladkaya [172]

Answer: Th enthalpy of combustion for the given reaction is 594.244 kJ/mol

Explanation: Enthalpy of combustion is defined as the decomposition of a substance in the presence of oxygen gas.

W are given a chemical reaction:

$Mg(s)+\frac{1}{2}O_2(g)\rightarrow MgO(s)$

$c=5760J/^oC$

$\Delta T=0.570^oC$

To calculate the enthalpy change, we use the formula:

$\Delta H=c\Delta T\\\\\Delta H=5760J/^oC\times 0.570^oC=3283.2J$

This is the amount of energy released when 0.1326 grams of sample was burned.

So, energy released when 1 gram of sample was burned is = $\frac{3283.2J}{0.1326g}=24760.181J/g$

Energy 1 mole of magnesium is being combusted, so to calculate the energy released when 1 mole of magnesium ( that is 24 g/mol of magnesium) is being combusted will be:

$\Delta H=24760.181J/g\times 24g/mol\\\\\Delta H=594244.3J/mol\\\\\Delta H=594.244kJ/mol$

4 0

2 years ago

A chemist pours 1 mol of zinc granules into one beaker and 1 mol of zinc chloride powder into another beaker. What do the two sa

irina1246 [14]

The number of Zn particles (atoms of Zn ) in the first sample, and Zn ions are going to be the same,
Also mass of Zn are going to be the same, because of electrons are too small, and would not have influence on mass.

6 0

2 years ago

Read 2 more answers

How much heat is required to raise the temperature of 20.0g of ice from-12°C to 0°C?

Vladimir79 [104]

Answer:

https://socratic.org/questions/how-much-heat-is-required-to-convert-5-88-g-of-ice-at-12-0-c-to-water-at-25-0-c-

Explanation:

6 0

2 years ago

3. Which of the following mass measurements is the smallest?

kogti [31]

It’s going to be ( A) because kg is the lowest measurement for the mass

8 0

2 years ago

What weight of sodium nitrate (NaNO3) must be used to prepare 506mL of a 10.5 M solution of this salt in water? Answer in units

topjm [15]

The question requires us to calculate the mass of NaNO3 necessary to prepare 506 mL of a 10.5 M solution of this salt.

To answer this question, we'll need to go through the following steps:

1) Calculate the molar mass of NaNO3, as it will be necessary during the question;

2) Calculate the number of moles necessary to prepare 506 mL of the 10.5 M solution;

3) Convert the number of moles calculated into the mass of NaNO3, using the molar mass of this salt.

Next, we'll follow the steps:

1) To calculate the molar mass of NaNO3, we need to consider the atomic masses of Na, N and O and the number of atoms of each of these elements. The atomic masses are 22.99 u, 14.01 u and 15.99 u for Na, N and O, respectively. With this information, we can calculate the molar mass:

molar mass of NaNO3 = (1 * 22.99) + (1 * 14.01) + (3 * 15.99) = 84.97 g/mol

2) Next, we use the molar concentration provided (10.5 M or 10.5 mol/L) and the volume required (506 mL = 0.506 L) to obtain the number of moles necessary to prepare this solution:

1 L of solution --------------- 10.5 mol of NaNO3

0.506 L of solution -------- x

Solving for x, we have:

$x=\frac{(0.506\text{ L of solution)}\times(10.5\text{ mol of NaNO}_3)}{(1\text{ L of solution)}}=5.31\text{ mol of NaNO}_3$

Therefore, the number of moles of NaNO3 necessary to prepare the solution is 5.31 moles.

3) At last, we use the molar mass obtained in step 1 (84.97 g/mol) to calculate the mass that corresponds to 5.31 moles of NaNO3:

1 mol of NaNO3 ---------------- 84.97 g

5.31 mol of NaNO3 ----------- y

Solving for y, we have:

$y=\frac{(5.31\text{ mol of NaNO}_3)\times(84.97\text{ g of NaNO}_3)}{(1\text{ mol of NaNO}_3)}=451.2\text{ g of NaNO}_3$

Therefore, the mass of NaNO3 necessary to prepare the solution given by the question is 451.2 g.

6 0

7 months ago