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prohojiy [21]
3 years ago
11

1) What is the mole ratio of D to A in the generic chemical reaction? (2 points) Page 390 helps with understanding mole ratios.

Be sure you’re only looking at the ratio of D to A. You can ignore B & C. 2A+ B  C + 3D
Chemistry
1 answer:
inna [77]3 years ago
4 0

Answer:- The mole ratio of D to A is 3:2.

Explanations:- Mole ratio for a chemical equation is the ratio of the coefficients of the molecules. Coefficients are the numbers used to balance the chemical equations.

For the given generic chemical equation, the coefficient of molecule A is 2 and the coefficient of D is 3. It means 2 molecules of A react with 3 molecules of D. So, the mole ratio of A to D is 2:3.

Since, we are asked about the mole ratio of D to A so we will write the coefficient of D first and hence the mole ratio of D to A is 3:2.

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Answer: 3.024 g grams of hydrogen are needed to  convert 76 grams of chromium(III) oxide, Cr_{2}O_{3}

Explanation:

The reaction equation for given reaction is as follows.

Cr_{2}O_{3} + 3H_{2} \rightarrow 2Cr + 3H_{2}O

Here, 1 mole of Cr_{2}O_{3} reacts with 3 moles of H_{2}.

As mass of chromium (III) oxide is given as 76 g and molar mass of chromium (III) oxide (Cr_{2}O_{3}) is 152 g/mol.

Number of moles is the mass of substance divided by its molar mass. So, moles of Cr_{2}O_{3} is calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{76 g}{152 g/mol}\\= 0.5 mol

Now, moles of H_{2}.given by 0.5 mol of Cr_{2}O_{3} is calculated as follows.

0.5 mol Cr_{2}O_{3} \times \frac{3 mol H_{2}}{1 mol Cr_{2}O_{3}}\\= 1.5 mol H_{2}

As molar mass of H_{2} is 2.016 g/mol. Therefore, mass of H_{2} is calculated as follows.

No. of moles = \frac{mass}{molar mass}\\1.5 mol = \frac{mass}{2.016 g/mol}\\mass = 3.024 g

Thus, we can conclude that 3.024 g grams of hydrogen are needed to  convert 76 grams of chromium(III) oxide, Cr_{2}O_{3}.

7 0
3 years ago
An unknown piece of metal weighing 95.0 g is heated to 98.0°C. It is dropped into 250.0 g of water at 23.0°C. When equilibrium i
lutik1710 [3]

Answer:

C_{metal}=126.6\frac{J}{g\°C}

Explanation:

Hello!

In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

Q_{metal}=-Q_{water}

Which can be also written as:

m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})

Thus, since we need the specific heat of the metal, we solve for it as shown below:

C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}

Best regards.

7 0
2 years ago
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