What is the mass of 13 gold atoms?

Explain why for static electricity to occur between two surfaces that are rubbed against each other, one of the materials has to

lianna [129]

The rubbing creates a negative charge that is carried by electrons. The electrons can build up t produce static electricity.

8 0

3 years ago

What evidence did dalton have that atoms actually exist?

likoan [24]

Dalton noted that the chromium oxide molecules that he was studying had masses of oxygen that were not the same: he found that there were constant ratios to these mass differences, however, so there had to be some sort of "units" (atoms) making up the ratios of these particles.

5 0

3 years ago

Oh, no! You just spilled 85.00 mL of 1.500 M sulfuric acid on your lab bench and need to clean it up immediately! Right next to

vredina [299]

Explanation:

We will balance equation which describes the reaction between sulfuric acid and sodium bicarbonate: as follows.

$H_2SO_4(aq) + 2NaHCO_3(s) \rightarrow Na2SO_4(aq) + 2H_2O(l) + 2CO_2(g)$

Next we will calculate how many moles of $H_2SO_4$ are present in 85.00 mL of 1.500 M sulfuric acid.

As, Molarity = $\frac{\text{moles of solute}}{\text{liters of solution
}}$

1.500 M = $\frac{n}{0.08500 L
}$

n = 0.1275 mol $H_2SO_4$

Now set up and solve a stoichiometric conversion from moles of $H_2SO_4$ to grams of $NaHCO_3$ . As, the molar mass of $NaHCO_3$ is 84.01 g/mol.

$0.1275 mol H_2SO_4 \times (\frac{2 mol NaHCO_3}{1 mol H_2SO_4}) \times (\frac{84.01 g NaHCO_3}{1 mol NaHCO_3})$

= 21.42 g $NaHCO_3$

So unfortunately, 15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.

4 0

3 years ago

When 10.0 grams of sulfur reacts with fluorine gas at a pressure of 2.69 atmosphere in a 5.00 L container at 0.00 degrees Celsiu

Gwar [14]

Answer:

74.1%

Explanation:

Based on the reaction:

S₈ + 16F₂ → 8SF₄

1 mole of sulfur reacts with 16 moles of F₂ to produce 8 moles of SF₄

To solve this question we must find the moles of each reactant in order to find the moles of SF₄. Thus, we can find the theoretical mass produced. Percent yield is:

Percent yield = Actual yield (25.0g) / Theoretical yield * 100

Moles S₈: 256.52g/mol

10.0g * (1mol / 256.52g) = 0.0390 moles

Moles F₂:

PV = nRT

PV/RT = n

Where P is pressure in atm, V is volume in liters, R is gas constant and T is absolute temperature (0°C = 273.15K)

2.69atm*5.00L / 0.082atmL/molK*273.15K = n

0.600 moles = n

For a complete reaction of 0.600 moles F₂ are required:

0.600mol F₂ * (1mol S₈ / 8 mol F₂) = 0.075 moles S₈

As there are just 0.0390 moles, S₈ is limiting reactant.

The theoretical moles and mass of SF₄ -Molar mass: 108.07g/mol- is:

0.0390 moles S₈ * (8mol SF₄ / 1mol S₈) = 0.312 moles SF₄ * (108.07g) =

33.7g

Percent yield = 25.0g / 33.7g * 100

= 74.1%

6 0

3 years ago

Which of the following naming rules would apply to CaCO3?

Alja [10]

Metal (Ca) Polyatomic anion (CO3). You do not need to use the roman numeral because Ca is in group 1. The answer is metal + polyatomic or Calcium Carbonate

3 0

3 years ago

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