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3 years ago

13

Determine the percent yield for the reaction between 28.1 g of Sb4O6 and excess C if 17.3 g of Sb is recovered along with an unk

own amount of CO

1 answer:

hammer [34]3 years ago

7 0

The answer is 73.71%. Hope I helped plus mark as brailiest ^-^

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How to go from a compound to a name and from a name to a compound?

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Convert between the chemical formula of an ionic compound and its name ... When naming ionic compounds, the cation retains the same name as the element.

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3 years ago

If a 95.0 gram sample of metal at 100.0 oC is placed in 50.0 g of water with an initial temperature of 22.5oC and the final temp

kaheart [24]

Answer : The specific heat of the metal is, $1.11J/g^oC$

Explanation

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of metal = 95.0g

$m_2$ = mass of water = 50.0 g

$T_f$ = final temperature of mixture = $48.5^oC$

$T_1$ = initial temperature of metal = $100.0.0^oC$

$T_2$ = initial temperature of water = $22.5^oC$

Now put all the given values in the above formula, we get

$(95.0g)\times c_1\times (48.5-100.0)^oC=-[(50.0g)\times 4.18J/g^oC\times (48.5-22.5)^oC]$

$c_1=1.11J/g^oC$

Therefore, the specific heat of the metal is, $1.11J/g^oC$

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3 years ago

In viewing a chemical equation, how do you determine the ratios that relate reactants and products?

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In viewing a chemical equation,we <span>determine the ratios that relate reactants and products </span>from the coefficients of the reactants and products .

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4 years ago

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Which crystalline solid would you expect to have the highest melting point?

Marianna [84]

Covalent network crystals usually have the highest melting points and boiling points.

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3 years ago

Calculate the mass of vanadium(V) oxide (V2O5) that contains a million (1.0 *10^6) vanadium atoms. Be sure your answer has a uni

sweet [91]

Answer : The mass of vanadium(V) oxide will be $1.51\times 10^{-16}g$

Explanation : Given,

Number of atoms of $V_2O_5$ = $1.0\times 10^{6}$

Molar mass of $V_2O_5$ = 181.88 g/mole

In $V_2O_5$ , there are 2 atoms of vanadium and 5 atoms of oxygen.

First we have to determine the moles of $V_2O_5$ .

As, $2\times 6.022\times 10^{23}$ number of vanadium atom present in 1 moles of $V_2O_5$

So, $1.0\times 10^{6}$ number of vanadium atom present in $\frac{1.0\times 10^{6}}{2\times 6.022\times 10^{23}}=8.3\times 10^{-19}$ moles of $V_2O_5$

Now we have to determine the mass of $V_2O_5$ .

$\text{Mass of }V_2O_5=\text{Moles of }V_2O_5\times \text{Molar mass of }V_2O_5$

$\text{Mass of }V_2O_5=(8.3\times 10^{-19}mole)\times (181.88g/mole)=1.51\times 10^{-16}g$

Therefore, the mass of vanadium(V) oxide will be $1.51\times 10^{-16}g$

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4 years ago

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