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Leviafan [203]
2 years ago
11

Need some help please?

Chemistry
1 answer:
VladimirAG [237]2 years ago
4 0
26. B. 28. B. 20. A.
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A calorimeter contains 100 g of water at 39.8 ºC. A 8.23 g object at 50 ºC is placed inside the calorimeter. When equilibrium ha
Tju [1.3M]
When equilibrium has been reached so, according to this formula we can get the specific heat of the unknown metal and from it, we can define the metal as each metal has its specific heat:

Mw*Cw*ΔTw = Mm*Cm*ΔTm

when 
Mw → mass of water
Cw → specific heat of water
ΔTw → difference in temperature for water 

Mm→ mass of metal
Cw→ specific heat of the metal
ΔTm → difference in temperature for metal

by substitution:

100g * 4.18 * (40-39.8) = 8.23 g * Cm * (50-40)

∴ Cm = 83.6 / 82.3 = 1.02 J/g.°C

when the Cm of the Magnesium ∴ the unknown metal is Mg
6 0
3 years ago
Determine the number of 3s electrons in Na .
Scorpion4ik [409]

Answer:

Explanation:The atomic number of sodium is 11. That is, the number of electrons in sodium is 11. Therefore, a sodium atom will have two electrons in the first shell, eight in the 2nd orbit, and an electron in the 3rd shell.

3 0
1 year ago
A chemical reaction that has a positive δg is best described as _____.
Ann [662]
A chemical reaction that has a positive g is best described as endergonic.
8 0
3 years ago
7.00 of Compound x with molecular formula C3H4 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25c.
beks73 [17]

Answer:

\Delta H_{f,C_3H_4}=276.8kJ/mol

Explanation:

Hello!

In this case, since the equation we use to model the heat exchange into the calorimeter and compute the heat of reaction is:

\Delta H_{rxn} =- m_wC_w\Delta T

We plug in the mass of water, temperature change and specific heat to obtain:

\Delta H_{rxn} =- (35000g)(4.184\frac{J}{g\°C} )(2.316\°C)\\\\\Delta H_{rxn}=-339.16kJ

Now, this enthalpy of reaction corresponds to the combustion of propyne:

C_3H_4+4O_2\rightarrow 3CO_2+2H_2O

Whose enthalpy change involves the enthalpies of formation of propyne, carbon dioxide and water, considering that of propyne is the target:

\Delta H_{rxn}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{f,C_3H_4}

However, the enthalpy of reaction should be expressed in kJ per moles of C3H4, so we divide by the appropriate moles in 7.00 g of this compound:

\Delta H_{rxn} =-339.16kJ*\frac{1}{7.00g}*\frac{40.06g}{1mol}=-1940.9kJ/mol

Now, we solve for the enthalpy of formation of C3H4 as shown below:

\Delta H_{f,C_3H_4}=3\Delta H_{f,CO_2}+2\Delta H_{f,H_2O}-\Delta H_{rxn}

So we plug in to obtain (enthalpies of formation of CO2 and H2O are found on NIST data base):

\Delta H_{f,C_3H_4}=3(-393.5kJ/mol)+2(-241.8kJ/mol)-(-1940.9kJ/mol)\\\\\Delta H_{f,C_3H_4}=276.8kJ/mol

Best regards!

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3 years ago
How many molecules of H2O are equivalent to 97.2g H2O
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como se denomina el proceso utilizado para descomponer el agua

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