What is the name of this compound

Calculate the molarity of 29.9 g of MgS in 911 mL of solution??

LUCKY_DIMON [66]

0.58 is the molarity of 29.9 g of MgS in 911 mL of solution.

Define the molarity of a solution.

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.

Given data:

Volume of solution = 911 mL=0.911L

Given the mass of MgS = 29.9 g

Molar mass of MgS = 56.38 g/mol

Moles of MgS = 0.5303299042

$Molality = \frac{Moles \;solute}{Volume \;of \;solution \;in \;litre}$

$Molality = \frac{0.5303299042}{0.911L}$

Molality = 0.5821403998=0.58

Hence, 0.58 is the molarity of 29.9 g of MgS in 911 mL of solution.

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3 0

2 years ago

Combustion analysis of 1.200 g of an unknown compound containing carbon, hydrogen, and oxygen produced 2.086 g of CO2 and 1.134

timurjin [86]

<span>the empirical formula is C3H8O2 You need to determine the relative number of moles of hydrogen and carbon. So you first calculate the molar mass of CO2 and H20 Atomic weight of carbon = 12.0107 Atomic weight of hydrogen = 1.00794 Atomic weight of oxygen = 15.999 Molar mass CO2 = 12.0107 + 2 * 15.999 = 44.0087 Molar mass H2O = 2 * 1.00794 + 15.999 = 18.01488 Now calculate the number of moles of CO2 and H2O you have Moles CO2 = 2.086 g / 44.0087 g/mole = 0.0474 mole Moles H2O = 1.134 g / 18.01488 g/mole = 0.062948 mole Calculate the number of moles of carbon and hydrogen you have. Since there's 1 carbon atom per CO2 molecule, the number of moles of carbon is the same as the number of moles of CO2. But since there's 2 hydrogen atoms per molecule of H2O, The number of moles of hydrogen is double the number of moles of H2O Moles Carbon = 0.0474 Moles Hydrogen = 0.062948 * 2 = 0.125896 Now we need to determine how much oxygen is in the compound. Just take the mass of the compound and subtract the mass of carbon and hydrogen. What's left will be the mass of oxygen. Then divide that mass by the atomic weight of oxygen to get the number of moles of oxygen we have. 1.200 - 0.0474 * 12.0107 - 0.125896 * 1.00794 = 0.503797 Moles oxygen = 0.503797 / 15.999 = 0.031489 So now we have a ratio of carbon:hydrogen:oxygen of 0.0474 : 0.125896 : 0.031489 We need to find a ratio of small integers that's close to that ratio. Start by dividing everything by 0.031489 (selected because it's the smallest value) getting 1.505288 : 3.998095 : 1 The 1 for oxygen and the 3.998095 for hydrogen look close enough. But the 1.505288 for carbon doesn't work. But it looks like if we double all the numbers, we'll get something close to an integer for everything. So do so. 3.010575 : 7.996189 : 2 Now this looks good. Rounding everything to an integer gives us 3 : 8 : 2 So the empirical formula is C3H8O2</span>

5 0

3 years ago

Hurry PLEASE HELP!

avanturin [10]

B. 11,540

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.

Usually radioactive elements have an unstable atomic nucleus.

General formulas used in decay:

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}$

T = duration of decay

t 1/2 = half-life

N₀ = the number of initial radioactive atoms

Nt = the number of radioactive atoms left after decaying during T time

Nt=25 g

No=100 g

t1/2=5770 years

$\tt 25=100\dfrac{1}{2}^{T/5770}\\\\\dfrac{1}{4}=\dfrac{1}{2}^{T/5770}\\\\2=T/5770\rightarrowT=11540~years$

7 0

3 years ago

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Cr2 ] = 0.892 M and [Fe2 ] = 0.0150

Effectus [21]

<u>Given:</u>

Concentration of Cr2+ = 0.892 M

Concentration of Fe2+ = 0.0150 M

<u>To determine:</u>

The cell potential, Ecell

<u>Explanation:</u>

The half cell reactions for the given cell are:

Anode: Oxidation

Cr(s) ↔ Cr2+(aq) + 2e⁻ E⁰ = -0.91 V

Cathode: Reduction

Fe2+ (aq) + 2e⁻ ↔ Fe (s) E⁰ = -0.44 V

------------------------------------------

Net reaction: Cr(s) + Fe2+(aq) ↔ Cr2+(aq) + Fe(s)

E°cell = E°cathode - E°anode = -0.44 - (-0.91) = 0.47 V

The cell potential can be deduced from the Nernst equation as follows:

Ecell = E°cell - (0.0591/n)log[Cr2+]/[Fe2+]

Here, n = number of electrons = 2

Ecell = 0.47 - 0.0591/2 * log[0.892]/[0.0150] = 0.418 V

Ans: The cell potential is 0.418 V

8 0

3 years ago

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Carbon monoxide (CO) gas reacts with oxygen (O2) gas to produce carbon dioxide (CO2) gas. If 1.00 L of carbon monoxide reacts wi

REY [17]

The volume is 1.0 L and the mass is 1.96 g

5 0

3 years ago

Read 2 more answers