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Vinvika [58]
3 years ago
10

How many moles of MgS2O3 are in 171 g of the compound?

Chemistry
1 answer:
Alexxx [7]3 years ago
8 0
Divide the 181g by the mol mass of the compound. Just add up the masses of the various elements.
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What can form as a result of a chemical reaction?
cestrela7 [59]

Answer is: new substance.

For example, synthesis chemical reaction: Ba + F₂ → BaF₂.

Synthesis reaction is a type of reaction in which multiple reactants combine to form a single product.

New substance, barium fluoride is formed, with different chemical and ohysical properties than reactants (barium and fluorine).

In barium fluoride, barium has oxidation number +2 and fluorine has oxidation number -1, so compound has neutral charge.

4 0
3 years ago
Read 2 more answers
Write a balanced equation for the combustion of C7H16(l) (heptane) -- i.e. its reaction with O2(g) forming the products CO2(g) a
JulsSmile [24]

Answer:

<u>The standard enthalpy of reaction = -4854.7kJ</u>

<u>The difference: </u>ΔH-ΔE = Δ(PV) = Δn.R.T = <u>9910.288 J ≈ 9.91 kJ</u>    

Explanation:

<u>The balanced chemical equation for the combustion of heptane</u>:

C₇H₁₆ (l) + 11 O₂ (g) → 7 CO₂ (g) + 8 H₂O (l)

Given: The standard enthalpy of formation (\Delta H _{f}^{\circ }) for: C₇H₁₆ (l) = -187.8 kJ/mol, O₂ (g) = 0 kJ/mol, CO₂ (g) = -393.5 kJ/mol, H₂O (l) = -286 kJ/mol

<u>To calculate the standard enthalpy of reaction (\Delta H _{r}^{\circ }) can be calculated by the Hess's law</u>:

\Delta H _{r}^{\circ } = \left [\sum \nu \cdot\Delta H _{f}^{\circ }(products)  \right ] - \left [\sum \nu\cdot\Delta H _{f}^{\circ }(reactants)  \right ]

Here, \nu is the stoichiometric coefficient

⇒ \Delta H _{r}^{\circ } =

\left [ 7\times \Delta H _{f}^{\circ }\left (CO_{2}\right )+ 8\times \Delta H _{f}^{\circ }\left (H_{2}O \right )\right ]

- \left [1\times \Delta H _{f}^{\circ }\left (C_{7}H_{16}\right ) +11\times \Delta H _{f}^{\circ }\left (O_{2} \right ) \right ]

=\left [ 7\times \left (-393.5 kJ/mol \right )+ 8\times \left (-286 kJ/mol \right )\right ]

-\left [1\times \left (-187.8 kJ/mol \right ) +11\times \left (0 kJ/mol \right ) \right ]

⇒ \Delta H _{r}^{\circ } = \left [ \left (-2754.5 \right )+ \left (-2288 \right )\right ]\left -[ \left (-187.8 \right ) +\left (0 \right )\right ]

⇒ \Delta H _{r}^{\circ } = \left [ -5042.5 ]\left -[ -187.8] = \left ( -4854.7kJ \right )

<u>To calculate the difference: </u>ΔH-ΔE=Δ(PV)

We use the ideal gas equation: P.V = n.R.T

⇒ ΔH-ΔE=Δ(PV) = Δn.R.T

Given: Temperature:T = 298K, R = 8.314 J⋅K⁻¹⋅mol⁻¹

Δn = number of moles of gaseous products - number of moles of gaseous reactants = (7)- (11) = (-4)

⇒ ΔH-ΔE=Δ(PV) = Δn.R.T = (-4 mol) × (8.314 J⋅K⁻¹⋅mol⁻¹) × (298K) = <u>9910.288 J = 9.91 kJ</u>                              (∵ 1 kJ = 1000J )

                                                                             

8 0
3 years ago
if a gas has an initial pressure of 24,650 pa and an initial volume of 376 ml, what is the final volume if the pressure of the g
NNADVOKAT [17]

If a gas has an initial pressure of 24,650 pa and an initial volume of 376 ml, then the final volume would be 11,943.8144 ml if the pressure of the gas is changed to 775 torr assuming that the amount and the temperature of the gas remain constant.

It is given that the initial pressure P₁ is 24,650Pa and initial volumeV₁ is 376ml and the final pressureP₂ is 775 torr. We need to find the final volume of the gas. The final volume could be found using the following formula:

P₁V₁ = P₂V₂

By substituting the values, we get

24650 x 376 = 776 x V₂

9268400 = 776V₂

V₂ = 9268400/776

V₂ = 11,943.8144 ml

Therefore, the final volume of the gas would be 11,943.8144 ml

To know more about Partial pressure, click below:

brainly.com/question/14119417

#SPJ4

5 0
1 year ago
explain how both hydrogen and carbon have achieved stability by bonding with each other to form methane ​
Naya [18.7K]

A carbon iota can bond with four other iotas and is just like the four-hole wheel, whereas an oxygen iota, which can bond only to two, is just like the two-hole wheel.

4 0
2 years ago
Read 2 more answers
The total pressure of gases a, b, and c in a closed container is 4.1 atm. if the mixture is 36% a, 42% b, and 22% c by volume, w
Natali [406]

 The  partial  pressure of  gas C  is 0.902  atm


  calculation

partial pressure of gas c  =[( percent by volume of  gas   C /  total  percent)   x total pressure]


percent  by  volume of gas C= 22%

Total   percent  = 36% +42%  + 22%  = 100 %

Total  pressure  =  4.1 atm


partial  pressure  of gas C  is therefore =  22/100 x 4.1 atm = 0.902  atm

3 0
3 years ago
Read 2 more answers
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