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3 years ago

How to calculate the percentage of a anhydrous salt

Chemistry

1 answer:

Anna35 [415]3 years ago

8 0

Yep u right bro but like you know I’m here with these guys you know what I mean tho

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2.20 moles Sn to grams

mihalych1998 [28]

261.162 grams. Use the equation n=M÷Mr.

3 0

2 years ago

Convert 8.50 moles Ca to atoms

Wewaii [24]

8.50 moles is equal to 5.1187×10²⁴ atoms of Ca.

Explanation:

We have to multiply the moles of Ca by the Avogadro's number:

= 6.022×10²³

So the number of atoms:

= 8.5 moles × 6.022×10²³atoms / mol

= 5.1187×10²⁴ atoms

Hence the 8.50 moles is equal to 5.1187×10²⁴ atoms of Ca.

4 0

3 years ago

Part a use these data to calculate the heat of hydrogenation of buta-1,3-diene to butane. c4h6(g)+2h2(g)→c4h10(g)

Reptile [31]

Answer: The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]$

For the given chemical reaction:

$C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]$

We are given:

$\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J$

Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

4 0

2 years ago

The following information is to be used for the next 2 questions. In order to analyze for Mg and Ca, a 24-hour urine sample was

Ainat [17]

Answer:

Explanation:

From the given information:

The concentration of metal ions are:

$[Ca^{2+}]= \dfrac{0.003474 \ M \times 20.49 \ mL}{10.0 \ mL}$

$[Ca^{2+}]=0.007118 \ M$

$[Mg^2+] = \dfrac{0.003474 \ M\times (26.23 - 20.49 )mL}{10.0 \ mL}$

$=0.001994 \ M$

Mass of Ca²⁺ in 2.00 L urine sample is:

$= 2.00 L \times 0.001994 \dfrac{mol}{L} \times \dfrac{40.08 \ g}{1 \ mol}$

= 0.1598 g

Mass of Ca²⁺ = 159.0 mg

Mass of Mg²⁺ in 2.00 L urine sample is:

$= 2.00 L \times 0.007118 \dfrac{mol}{L} \times \dfrac{24.31 \ g}{1 \ mol}$

= 0.3461 g

Mass of Mg²⁺ = 346.1 mg

5 0

3 years ago

Consider a balloon with volume V. It contains n moles of gas and has an internal pressure of P. The temperature of the gas is T.

Maksim231197 [3]

Answer:

V₂ = 0.6 V.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n is constant, and have different values of P, V and T:

(P₁V₁T₂) = (P₂V₂T₁).

V₁ = V, P₁ = P, T₁ = T.

V₂ = ??? V, P₂ = 1.25 P, T₂ = 0.75 T.

∴ V₂ = (P₁V₁T₂)/(P₂T₁) = (P)(V)(0.75 T)/(1.25 P)(T) = 0.6 V.

3 0

3 years ago

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