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Zina [86]
3 years ago
13

1.00-degree increase on the Celsius scale is equivalent to a 1.80-degree increase on the Fahrenheit scale. The temperature of a

fluid increases by 48.0°C. What is this increase in degrees Fahrenheit? 49.8°F 48.0°F 26.7°F 86.4°F
Chemistry
1 answer:
castortr0y [4]3 years ago
6 0
48.0°C×1.80 = 86.4°F
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Cryolite, Na3AlF6(s), an ore used in the production of aluminum, can be synthesized using aluminum oxide. Balance the equation f
Korvikt [17]

Answer:

55.2kgNa_{3}AlF_{6}

Explanation:

1. First balance the equation for the synthesis of cryolite:

Al_{2}O_{3}_{(s)}+6NaOH_{(l)}+12HF_{(g)}=2Na_{3}AlF_{6}+9H_{2}O_{(g)}

2. Find the limiting reagent between the Al_{2}O_{3},NaOH and HF

- First calculate the number of moles of each compound using its molar mass and the mass that reacted completely:

13.4kgAl_{2}O_{3}*\frac{1molAl_{2}O_{3}}{101.96gAl_{2}O_{3}}*\frac{1000g}{1kg}=131molesAl_{2}O_{3}

55.4kgNaOH*\frac{1molNaOH}{40kgNaOH}*\frac{1000g}{1kg}=1385molesNaOH55.4kgHF*\frac{1molHF}{20kgHF}*\frac{1000g}{1kg}=2770molesHF

- Divide the number of moles obtained between the stoichiometric coefficient of each compound in the chemical reaction:

Al_{2}O_{3}:\frac{131}{1}=131

NaOH:\frac{1385}{6}=231

HF:\frac{2770}{12}=231

The Al_{2}O_{3} is the limiting reagent because it has the smallest number.

3. Find the mass of cryolite produced:

13.4kgAl_{2}O_{3}*\frac{1molAl_{2}O_{3}}{0.10196kgAl_{2}O_{3}}*\frac{2molesNa_{3}AlF_{6}}{1molAl_{2}O_{3}}*\frac{0.20994kgNa_{3}AlF_{6}}{1molNa_{3}AlF_{6}}=55.2kgNa_{3}AlF_{6}

3 0
4 years ago
Calculate the pOH for a solution with a hydroxide ion, (OH-) of concentration of 6.49 X 10-11 M.
Ray Of Light [21]

Answer:

Calculate the pOH for a solution with a hydroxide ion, (OH-) of concentration of 6.49 X 10-11 M.

Calculate the pH.

Note: the answers should have three significant figures

The pOH is: Blank 1

The pH is: Blank 2

Explanation:

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Oxidation number for O in: AIPO4
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Answer:

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Explanation:

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