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3 years ago

A student gains 1.0 lb in two weeks. How many grams did he gain?

2 answers:

7 0

1lb= 453.592g but then to change to 2 significant figures you go down two of the numbers so here it will be 4 and 5 that stands for 450g. So he gained 450g. I hope that helped

OlgaM077 [116]3 years ago

3 0

1 lb=453.592 grams
So he gained 453.592 pounds

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The burning of magnesium is a highly exothermic reaction. How many kilojoules of heat are released when 0. 75 mol of Mg burn in

alexandr402 [8]

Taking into account the definition of enthalpy of a chemical reaction, the quantity of heat released when 0.75 moles of Mg are burned is 451.5 kJ.

<h3>Enthalpy of a chemical reaction</h3>

The enthalpy of a chemical reaction is known as the heat absorbed or released in a chemical reaction when it occurs at constant pressure. That is, the heat of reaction is the energy that is released or absorbed when chemicals are transformed into a chemical reaction.

The enthalpy is an extensive property, that is, it depends on the amount of matter present.

<h3>Heat released in this case</h3>

In this case, the balanced reaction is:

2 Mg(s) + O₂ (g) → 2 MgO(s) + 1204 kJ

This equation indicates that when 2 moles of Mg reacts with 1 mole of O₂, 1204 kJ of heat is released.

When 0.75 moles of Mg are burned, then you can apply the following rule of three: if 2 moles of Mg releases 1204 kJ of heat, 0.75 moles of Mg releases how much heat?

$heat=\frac{0.75 moles of Mgx1204 kJ}{2 moles of Mg}$

Finally, the quantity of heat released when 0.75 moles of Mg are burned is 451.5 kJ.

Learn more about enthalpy of a chemical reaction:

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4 0

2 years ago

A compound contains sulfur, oxygen, and chlorine. Analysis shows that it contains by mass 26.95% sulfur and 59.61% chlorine. Wha

sladkih [1.3K]

Answer:

SCl₂

Explanation:

In order to know the empirical formula, we have to follow a series of steps.

Step 1: Divide each percentage by the atomic mass

S: 26.95/32.07 = 0.8403

Cl: 59.61/35.45 = 1.682

Step 2: Divide all the numbers by the smallest one.

S: 0.8403/0.8403 = 1

Cl: 1.682/0.8403 ≈ 2

The empirical formula of the compound is SCl₂.

7 0

3 years ago

Read 2 more answers

Consider two gases, A and B, are in a container at room temperature. What effect will the following changes have on the rate of

Tcecarenko [31]

Answer:

decrease in temperature , decreases the kinetic movements of the gase molecules as a result decreases the frequency of collisions between gas molecules A and B consequently decreases the rate of reactions of gases A and B

6 0

2 years ago

Assuming that gasoline is 100% isooctane, that isooctane burns to produce only CO2CO2 and H2OH2O, and that the density of isooct

Aleksandr [31]

Answer:

1.12×10¹¹ kg of CO₂ are produced with 4.6×10¹⁰ L of isooctane

Explanation:

Let's state the combustion reaction:

C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O

Let's calculate the mass of isooctane that reacts.

Density = Mass / Volume

Density . Volume = Mass

First of all, let's convert the volume in L to mL, so we can use density.

4.6×10¹⁰ L . 1000 mL / 1L = 4.6×10¹³ mL

0.792 g/mL . 4.6×10¹³ mL = 3.64 ×10¹³ g

This mass of isooctane reacts to produce CO₂ and water, so let's determine the moles of reaction

3.64 ×10¹³ g . 1mol / 114 g = 3.19×10¹¹ mol

Ratio is 1:8 so 1 mol of isooctane can produce 8 moles of dioxide

Therefore 3.19×10¹¹ mol would produce (3.19×10¹¹ mol . 8) = 2.55×10¹² moles of CO₂

Now, we can determine the mass of produced CO₂ by multipling:

moles . molar mass

2.55×10¹² mol . 44 g/mol = 1.12×10¹⁴ g of CO₂

If we convert to kg 1.12×10¹⁴ g / 1000 = 1.12×10¹¹ kg

6 0

3 years ago

Cans of regular soda have volumes with a mean of 14.5 oz and a standard deviation of 0.18 oz. What is the maximum "usual" value

tangare [24]

Answer:

14.68 oz

Explanation:

The mean volume is the average volume of all the regular soda cans put together; 14.5 oz. The standard deviation 0.18 oz is the maximum deviation from the mean. It can be 14.5-0.18= 14.32 oz which is the usual minimum or 14.5 + 0.18=14.68 oz which is the usual maximum

4 0

3 years ago