10 Points! Which type of fossil is formed when a rock hardens inside a mold fossil?

A 25 kg rock is placed in a graduated cylinder with water.the volume of the fluid is 18.3ml.calculate the density of the rock in

ipn [44]

Answer:

=> 1366.120 g/mL.

Explanation:

To determine the formula to use in solving such a problem, you have to consider what you have been given.

We have;

mass (m) = 25 Kg

Volume (v) = 18.3 mL.

From our question, we are to determine the density (rho) of the rock.

The formula:

$p = \frac{m}{v}$

First let's convert 25 Kg to g;

1 Kg = 1000 g

25 Kg = ?

$= \frac{25 × 1000}{1}$

= 25000 g

Substitute the values into the formula:

$p = \frac{25000 g}{18.3 ml}$

= 1366.120 g/mL.

Therefore, the density (rho) of the rock is 1366.120 g/mL.

8 0

2 years ago

A sample of a compound contains 3.21 g of sulfur and 11.4 g of fluorine. Which of the following represents the empirical formula

Sergeeva-Olga [200]

Answer:

The empirical formula is SF6 (option E)

Explanation:

Step 1: Data given

Mass of sulfur = 3.21 grams

Mass of fluorine = 11.4 grams

Molar mass sulfur = 32.065 g/mol

Molar mass fluorine = 19.00 g/mol

Step 2: Calculate moles

Moles = mass /molar mass

Moles sulfur = 3.21 grams / 32.065 g/mol

Moles sulfur = 0.100 moles

Moles fluorine = 11.4 grams / 19.00 g/mol

Moles fluorine = 0.600 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

S: 0.100 / 0.100 = 1

F : 0.600 / 0.100 = 6

The empirical formula is SF6 (option E)

7 0

3 years ago

When 100 mL of 0.200 M NaCl(aq) and 100 mL of 0.200 M AgNO3(aq), both at 21.9 °C, are mixed in a coffee cup calorimeter, the tem

masya89 [10]

Answer:

There is 1.3 kJ heat produced(released)

Explanation:

Step 1: Data given

Volume of a 0.200 M Nacl solution = 100 mL = 0.1 L

Volume of a 0.200 M AgNO3 solution = 100 mL = 0.1 L

Initial temperature = 21.9 °C

Final temperature = 23.5 °C

Solid AgCl will be formed

Step 2: The balanced equation:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

AgCl(s) + NaNO3(aq) → Na+(aq) + NO3-(aq) + AgCl(s)

Step 3: Define the formula

Pressure is constant. → the heat evolved from the reaction is equivalent to the enthalpy of reaction.

Q=m*c*ΔT

⇒ Q = the heat transfer (in joule)

⇒ m =the mass (in grams)

⇒ c= the heat capacity (J/g°C)

⇒ ΔT = Change in temperature = T2- T1

Step 4: Calculate heat

Let's vonsider the density the same as the density of water (1g/mL)

Mass = volume * density

Mass = 200 mL * 1g/mL

Mass = 200 grams

Q= m*c*ΔT

⇒ m = 200 grams

⇒ c = the heat capacity (let's consider the heat capacity of water) = 4.184 J/g°C

⇒ ΔT = 23.5 -21.9 = 1.6°C

Q = 200 * 4.184 * 1.6 = 1338 .9 J = 1.3 kJ

There is 1.3 kJ heat produced(released)

Therefore, we assumed no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and the specific heat and mass of the solution are the same as those for water (1g/mL and 4.184 J/g°C)

7 0

3 years ago

A sample of 3.62 moles of diphosphorous trioxide is

Sindrei [870]

$H_3PO_3$ $2H_3PO_3$ $2H_3PO_3$ $P_2O_3$ Answer:

B

Explanation:

This question is about stoichiometry. From the balanced equation $P_2O_3 + 3H_2O$ ⇒ $2H_3PO_3$ , we see that 3 moles of water is needed to react with 1 mole of $P_2O_3$ .

This means that, to fully react 3.62 moles of $P_2O_3$ , we would need 3*3.62 or 10.86 moles of water. However, we only have 6.31 moles, so water is the limiting reactant.

Since 3 moles of water react with 1 mole of $P_2O_3$ , 6.31 moles of water can fully react with 6.31÷3 or 2.1033 moles of $P_2O_3$ .

From the balanced equation, we see that every mole of $P_2O_3$ reacted gets you 2 moles of $2H_3PO_3$ . Therefore, 2.1033 moles of $P_2O_3$ would give you approximately 4.21 moles of $H_3PO_3$ .

5 0

2 years ago

Is photosphere the layer that emits sunlight?

weqwewe [10]

Answer:

In our Sun, as in other stars, roughly 99.9% or so of all light emitted is emitted in a thin layer known as the photosphere, or light sphere. This is explained as follows. Interior to the photosphere the gas is ever denser and becomes far too opaque for any photon to emerge directly from that layer.

Explanation:

4 0

3 years ago

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