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3 years ago

Which radioisotope is a beta emitter?(1) 90Sr (3) 37K(2) 220Fr (4) 238U

1 answer:

4 0

The answer is (1) 90 Sr. The definition of beta decay is that a type of radioactive decay in which one proton will change to one neutron and release one electron inside the nucleus. And 90Sr is a common beta emitter.

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Earthquakes , volcanoes mountains and oceans trenches are most closely associated with which plate boundary?

masya89 [10]

Answer:

continental plate boundary

8 0

3 years ago

Why do igneous rocks not contain fossils in them ?

Mashutka [201]

Answer:

Igneous rocks do not contain any fossils.

Explanation:

This is because any fossils in the original rock will have melted when the rock melted to form magma. Sedimentary rocks are formed from the broken remains of other rocks that become joined together.

6 0

3 years ago

Help please please please please

Rom4ik [11]

Answer:

D. 18.7 grams

Explanation:

The coefficients are the key.

Create a proportion with them, and the molar mass, and then solve for x:

$\frac{2}{3} = \frac{x}{28} \\ 3x = 2(28) \\ 3x = 56 \\ x = 18.666... = 18.7$

7 0

3 years ago

A gas is initially confined to a 1.00 L vessel at 3.00 atm of pressure. A valve connecting the 1.00 L vessel to a 3.00 L vessel

Brilliant_brown [7]

Answer:

c. 0.750 atm .

Explanation:

Hello!

In this case, since the two vessels have different volume, we can see that the gas is initially at 3.00 atm into the 1.00-L vessel, but next, it is allowed to move towards the 3.00-L vessel, meaning that the final volume wherein the gas is located, is 4.00 L; therefore, we use the Boyle's law to compute the final pressure:

$P_2V_2=P_1V_1\\\\P_2=\frac{P_1V_1}{V_2} \\\\P_2=\frac{3.00atm*1.00L}{4.00L}\\\\P_2=0.750atm$

Therefore the answer is c. 0.750 atm .

Best regards!

8 0

3 years ago

The complete combustion of propane (C3H8) in the presence of oxygen yields CO2 and H2O: C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (g) a

mixas84 [53]

Answer:

26.9 L is the volume of CO₂, we obtained

Explanation:

The reaction is: C₃H₈(g) + 5O₂(g) → 3CO₂ (g) + 4H₂O (g)

Let's determine the reactants moles:

27.5 g . 1mol / 44 g = 0.625 moles

We need density of O₂ to determine mass and then, the moles.

O₂ density = O₂ mass / O₂ volume

O₂ density . O₂ volume = O₂ mass

1.429 g/L . 45L = O₂ mass → 64.3 g

Moles of O₂ → 64.3 g . 1mol/32g = 2.009 moles

Let's find out the limiting reactant:

1 mol of propane needs 5 moles of oxygen to react

Then, 0.625 moles will react with (0.625 . 5)/1 = 3.125 moles of O₂

Oxygen is the limiting reactant, we need 3.125 moles but we only have 2.009 moles

Ratio is 5:3. 5 moles of O₂ produce 3 moles of CO₂

Therefore, 2.009 moles of O₂ must produce (2.009 .3) /5 = 1.21 moles of CO₂. Let's find out the volume, by Ideal Gases Law (STP are 1 atm and 273K, the standard conditions)

1 atm . V = 1.21 moles . 0.082 . 273K

V = (1.21 moles . 0.082 . 273K) / 1atm = 26.9 L

3 0

3 years ago