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2 years ago

Changing the number of element produces isotopes of the same

Chemistry

1 answer:

Ber [7]2 years ago

3 0

Answer:

An isotope is one of two or more forms of the same chemical element. Different isotopes of an element have the same number of protons in the nucleus, giving them the same atomic number, but a different number of neutrons giving each elemental isotope a different atomic weight.

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Hydrogen gas explosive within range of 4%-75% v/v. Assuming that each student in your class produces 6 L of H2 and that the lab

SpyIntel [72]

To determine whether the amount of H2 in the lab is dangerous, we first need to know how much hydrogen gas is present in the room in units of percent by volume. For this particular problem, we cannot exactly determine since we do not know the total volume of the room. Hope this answers the question.

3 0

4 years ago

a 25.0-ml volume of a sodium hydroxide solution requires 19.6 ml of a 0.189 m hydrochloric acid for neutralization. a 10.0- ml v

Rashid [163]

Concentration of NaOH = 0.148 molar, M

Concentration of H3PO4 = 0.172 molar, M

Concentration x Volume will give the number of moles of solute in that volume. C*V = moles

Concentration has a unit of (moles/liter). When multiplied by the liters of solution used, the result is the number of moles.

Original HCl solution: (0.189 moles/L)*(0.0196 L)= 0.00370 moles of HCl

The neutralization of 25.0 ml of sodium hydroxide, NaOH, requires 0.00370 moles of HCl. The reaction is:

NaOH + HCl > NaCl and H2O

This balanced equation tells us that neutralization of NaOH with HCl requires the same number of moles of each. We just determined that the moles of HCl used was 0.00370 moles. Therefore, the 25.0 ml solution of NaOH had the same number of moles: 0.00370 moles NaOH.

The 0.00370 moles of NaOH was contained in 25.0 ml (0.025 liters). The concentration of NaOH is therefore:

(0.00370 moles of NaOH)/(0.025 L) = 0.148 moles/liter or Molar, M

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The phosphoric acid problem is handled the same way, but with an added twist. Phosphoric acid is H3PO4. We learn the 34.9 ml of the same NaOH solution (0.148M) is needed to neutralize the H3PO4. But now the acid has three hydrogens that will react. The balanced equation for this reaction is:

H3PO4 + 3NaOH = Na3PO4 + 3H2O

Now we need three times the moles of NaOH to neutralize 1 mole of H3PO4.

The moles of NaOH that were used is:

(0.148M)*(0.0349 liters) = 0.00517 moles of NaOH

Since the molar ratio of NaOH to H3PO4 is 3 for neutralization, the NaOH only neutralized (0.00517)*(1/3)moles of H3PO4 = 0.00172 moles of H3PO4.

The 0.00172 moles of H3PO4 was contained in 10.0 ml. The concentration is therefore:

(0.00172 moles H3PO4)/(0.010 liters H3PO4)

Concentration of H3PO4 = 0.172 molar, M

5 0

1 year ago

How do you solve for density

vodka [1.7K]

The Density Calculator uses the formula p=m/V, or density (p) is equal to mass (m) divided by volume (V). The calculator can use any two of the values to calculate the third. Density is defined as mass per unit volume.

5 0

3 years ago

Read 2 more answers

The molar heat capacity of ethane is represented in the temperature range 298 K to 400 K by the empirical expression Cp,m in J K

BabaBlast [244]

Answer:

-88.66 kJ/mol

Explanation:

The expressions of heat capacity (Cp,m) for C(s) and for H₂(g) are:

C(s): Cp,m/(J K-1 mol-1) = 16.86 + (4.77T/10³) - (8.54x10⁵/T²)

H₂(g): Cp,m/(J K-1 mol-1) = 27.28 + (3.26T/10³) + (0.50x10⁵/T²)

Cp = A + BT + CT⁻²

For the Kirchoff's Law:

ΔHf = ΔH°f + $\int\limits^{T2}_{T1} {DCp(T)} \, dT$

Where ΔH°f is the enthalpy at 298 K, T1 is 298 K, T2 is the temperature given (373 K), and DCp is the variation of Cp (products less reactants). ΔH°f for ethene is -84.68 kJ/mol and the reaction is:

2C(s) + 3H₂(g) → C₂H₆

So, DCp:

dA = A(C₂H₆) - [2xA(C) + 3xA(H₂)] = 14.73 - [2x16.86 + 3x27.28] = -100.83

dB = B(C₂H₆) - [2xB(C) + 3xB(H₂)] = 0.1272 - [2x4.77x10⁻³ + 3x3.26x10⁻³] = 0.10788

dC = C(C₂H₆) - [2xC(C) + 3xC(H₂)] = 0 - (2x(-8.54x10⁵) + 3x0.50x10⁵) = 15.58x10⁵

dCp = -100.83 + 0.10788T + 15.58x10⁵T⁻²

$\int\limits^{373}_{298} {-100.83 + 0.10788T + 15.58x10^5T^{-2}} \, dT$ = -3796.48 J/mol = -3.80 kJ/mol (solved by a graphic calculator)

ΔHf = -84.68 - 3.80

ΔHf = -88.66 kJ/mol

7 0

3 years ago

A student measures the pressure and volume of an empty water bottle to be 1. 4 atm and 2. 3 l

rewona [7]

The new volume of the bottle is 5.07 L.

<h3>What is Boyle's Law?</h3>

It is a gas law that states the pressure decrease with the increase in the pressure.

By the formula

$\rm P_1V_1= P_1V_2$

Given that, volume1 is 2.3l

Pressure1 is 4 atm.

Pressure2 to 0.65 atm.

V2 is to find?

Putting the values in the equation

$\rm 1.4\times 2.3 = 0.65 \times V_2\\\\V_2 = 5.07 L$

Thus, the new volume of the bottle is 5.07 L.

Learn more about Boyle's Law

brainly.com/question/1437490

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6 0

2 years ago