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3 years ago

How many covalent bonds are there in one molecule of carbon dioxide, CO2?

Chemistry

2 answers:

Nutka1998 [239]3 years ago

6 0

It would be two because there’s 1 Carbon and 2 oxygen

ikadub [295]3 years ago

3 0

Answer:

four

Explanation:

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GarryVolchara [31]

Answer:

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3 years ago

0.00099989 3 sig figs, scientific notation

mixas84 [53]

1.00*10^3
You’d need to lower the exponent because rounding to 3 sig figs changes the 9’s to - 1000. Keep the 0’s.

5 0

3 years ago

How many grams of hydrogen in 1.85moles

olga2289 [7]

2H2(g) + O2(g) → 2H2O(1) 0 260 g 0.2068 0.180 g 2008 When 45.0 g of CH4 reacts with excess O2, the actual yield of CO2 is 118 g. What is the percent yield? CHA(g) + 2O2(g) - CO2(g) + 2H2O(g) 73.6% 67.9% 95.2% 86.4% For the reaction: 2503(g) + 790 kcal - 25(s) + 3O2(g), how many kcal are needed to form 1.5 moles O2(g)? 790 kcal 395 kcal 2370 kcal 411 kcal When 3 moles of Ny are mixed with 5 moles of H2 the limiting reactant is N2(g) + 3H2(g) - 2NH3(g) H2 NH3 ОООО H20 O N₂

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3 years ago

Which sediment type requires the least amount of energy to be eroded

castortr0y [4]

Answer:

SEDIMENTARY ROCK

Explanation:

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3 years ago

In the laboratory a student combines 26.2 mL of a 0.234 M chromium(III) acetate solution with 10.7 mL of a 0.461 M chromium(III)

Natalka [10]

Answer: The molarity of $Cr^{3+}$ ions in the solution is 0.299 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For chromium (III) acetate:

Molarity of chromium (III) acetate solution = 0.234 M

Volume of solution = 26.2 mL

Putting values in equation 1, we get:

$0.234=\frac{\text{Moles of chromium (III) acetate}\times 1000}{26.2}\\\\\text{Moles of chromium (III) acetate}=\frac{0.234\times 26.2}{1000}=0.00613mol$

1 mole of chromium (III) acetate $(Cr(CH_3COO)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of acetate $(CH_3COO^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00613)=0.00613moles$

For chromium (III) nitrate:

Molarity of chromium (III) nitrate solution = 0.461 M

Volume of solution = 10.7 mL

Putting values in equation 1, we get:

$0.461=\frac{\text{Moles of chromium (III) nitrate}\times 1000}{10.7}\\\\\text{Moles of chromium (III) nitrate}=\frac{0.461\times 10.7}{1000}=0.00493mol$

1 mole of chromium (III) nitrate $(Cr(NO_3)_3)$ produces 1 mole of chromium $(Cr^{3+})$ ions and 3 moles of nitrate $(NO_3^-)$ ions

Moles of $Cr^{3+}\text{ ions}=(1\times 0.00493)=0.00493moles$

For chromium cation:

Total moles of chromium cations = [0.00613 + 0.00493] = 0.01106 moles

Total volume of solution = [26.2 + 10.7] = 36.9 mL

Putting values in equation 1, we get:

$\text{Molarity of }Cr^{3+}\text{ cations}=\frac{0.01106\times 1000}{36.9}\\\\\text{Molarity of }Cr^{3+}\text{ cations}=0.299M/tex]Hence, the molarity of [tex]Cr^{3+}$ ions in the solution is 0.299 M

5 0

3 years ago