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wel
3 years ago
14

Write the chemical formulas for the ionic compound vanadium (v) oxide

Chemistry
1 answer:
olya-2409 [2.1K]3 years ago
5 0
V2O5 is the formula. Vanadium has a charge of +5 and oxygen is -2, so to make them equal, you need to double the vanadium charge and multiply the oxygen charge by 5
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Convert a speed of 259 cm/s to units of inches per minute. Also, show the unit analysis by dragging components into the unit‑fac
igomit [66]

Answer:

6118.11\frac{in}{min}

Explanation:

Hello,

In this case, by knowing that 1 inch equals 2.54 cm and 60 seconds equals 1 min, the resulting value results:

=259\frac{cm}{s}*\frac{1in}{2.54cm}*\frac{60s}{1min}\\   \\=6118.11\frac{in}{min}

Best regards.

5 0
3 years ago
Calculate the atomic mass of chromium it’s composition is 83.79% with a mass of 51.94 amu; 9.50% with a mass of 52.94 Amu; 4.35%
Elina [12.6K]

<span><span>Convert the percentages into decimals (you can do that by dividing the percent by 100), then multiply that by its corresponding mass to find its relative amount/ contribution to the atomic mass of chromium. After doing so, add all of the obtained values together to get the average mass.

</span> 83.79% = .08379
9.50% = .095
4.35% = .0435
2.36% = .0236

Average mass of chromium = 0.8379(51.94) + 0.095( 52.94) + 0.0435(49.95) + 0.0236(53.94)

Answer: 52amu

P.S. never forget units
</span>
8 0
3 years ago
Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given
tekilochka [14]

Answer:

a) = 0.704%

b) = 1.30%

c) = 2.60%

Explanation:

Given that:

K_a= 1.34*10^{-5

For Part A; where Concentration of A = 0.270 M

Percentage Ionization(∝)  \alpha = \sqrt{\frac{K_a}{C} }

\alpha = \sqrt{\frac{1.34*10^{-5}}{0.270} }

\alpha = \sqrt{4.9629*10^{-5}}

\alpha = 7.044*10^{-3

percentage% (∝) = 7.044*10^{-3}*100

= 0.704%

For Part B; where Concentration of B = 7.84*10^{-2 M

\alpha = \sqrt{\frac{1.34*10^{-5}}{7.84*10^{-2}} }

\alpha = \sqrt{1.709*10^{-4} }

\alpha = 0.0130\\

percentage% (∝) = 0.0130 × 100%

= 1.30%

For Part C; where Concentration of C= 1.92*10^{-2} M

\alpha = \sqrt{\frac{1.34*10^{-5}}{1.97*10^{-2}} }

\alpha = \sqrt{6.802*10^{-4}}

\alpha =0.02608

percentage% (∝) = 0.02608  × 100%

= 2.60%

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3 years ago
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How many grams of magnesium oxide are needed to produce 264g of magnesium hydroxide
Archy [21]
If its one part magnesium and two parts hydroxide id say 88g of magnesium
3 0
3 years ago
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