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3 years ago

Isotopes of the same element must have the same number of - a protons + neutrons. b neutrons. c electrons. d protons.

2 answers:

4 0

Your answer is A....They have the same number of protons and neutrons, but a different number of electrons. They have the same number of protons and electrons, but a different number of neutrons. Different isotopes of an element have the same atomic number, but different mass numbers.....

scZoUnD [109]3 years ago

3 0

D. Protons
Isotopes of the same element share the same amount of protons only. The number of neutrons can be determined depending on the molar mass of the element minus the number of protons. Example: Isotopes of Carbon are Carbon 12 and Carbon 14
Carbon 12 has a molar mass ( molecular weight) of 12 minus 6 protons it has will give you 6 neutrons.
Carbon 14 will have (14-6) = 8 neutrons.
The number of electrons will be the same in both elements as long as they remain as neutral elements

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iris [78.8K]

The answer is elements

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3 years ago

3. The following data of decomposition reaction of thionyl chloride (SO2Cl2) were collected at a certain temperature and the con

KonstantinChe [14]

Answer:

a) First-order.

b) 0.013 min⁻¹

c) 53.3 min.

d) 0.0142M

Explanation:

Hello,

In this case, on the attached document, we can notice the corresponding plot for each possible order of reaction. Thus, we should remember that in zeroth-order we plot the concentration of the reactant (SO2Cl2 ) versus the time, in first-order the natural logarithm of the concentration of the reactant (SO2Cl2 ) versus the time and in second-order reactions the inverse of the concentration of the reactant (SO2Cl2 ) versus the time.

a) In such a way, we realize the best fit is exhibited by the first-order model which shows a straight line (R=1) which has a slope of -0.0013 and an intercept of -2.3025 (natural logarithm of 0.1 which corresponds to the initial concentration). Therefore, the reaction has a first-order kinetics.

b) Since the slope is -0.0013 (take two random values), the rate constant is 0.013 min⁻¹:

$m=\frac{ln(0.0768)-ln(0.0876)}{200min-100min} =-0.0013min^{-1}$

c) Half life for first-order kinetics is computed by:

$t_{1/2}=\frac{ln(2)}{k}=\frac{ln(2)}{0.013min^{-1}} =53.3min$

d) Here, we compute the concentration via the integrated rate law once 1500 minutes have passed:

$C=C_0exp(-kt)=0.1Mexp(-0.013min^{-1}*1500min)\\\\C=0.0142M$

Best regards.

6 0

3 years ago

Why are large differences in temperature between the wet bulb and dry bulb thermometers are possible only at higher temperatures

photoshop1234 [79]

Answer:

At constant vapor pressure, the relative humidity decreases as the temperature increases, therefore, at higher temperature the relative humidity is low and water readily evaporates from the wet bulb thermometer that results in the cooling of the bulb such that at a given ambient temperature the very low relative humidity results in very large differences between the temperatures of the wet bulb thermometer and that of the dry bulb thermometer and the wet bulb is observed to be the colder thermometer of the two

Explanation:

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3 years ago

URGENT PLEASE HELP!! Earth science!!

aev [14]

Answer:

URGENT PLEASE HELP!! Earth science!!

What stages of development are represented by the Group 2 and Group 3 stars? Describe these

groups in terms of their relative age, size, brightness, and temperature.

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3 years ago

I NEED HELP PLEASE, THANKS! :)

krok68 [10]

Answer:

$\large \boxed{\text{2.20 g Pb}}$

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 239.27 32.00 207.2

2PbS + 3O₂ ⟶ 2Pb + 2SO₃

m/g: 2.54 1.88

2. Calculate the moles of each reactant

$\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}$

3. Calculate the moles of Pb from each reactant

$\textbf{From PbS:}\\\text{Moles of Pb} = \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}$

4. Calculate the mass of Pb

$\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}$

5 0

3 years ago