Why is current more dangerous than voltage?

2 answers:

8 0

. The higher the current, the more likely it is lethal. Since current is proportional to voltage when resistance is fixed (ohm's law), high voltage is an indirect risk for producing higher currents.

Serga [27]3 years ago

6 0

The higher the current, the more likely it is lethal. Since current is proportional to voltage when resistance is fixed (ohm's law), high voltage is an indirect risk for producing higher currents.

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Calculate the mass of 3.4 moles of nitric acid (hno3). explain the process or show your work by including all values used to det

balu736 [363]

Answer:

6.34917360^25g

Explanation:

It's been a while since I've done this type of problem so I'm not making any promises that its right hahaha, but I hope it helps anyway. Please let me know whether I'm right or not!

6 0

1 year ago

Calculate the total pressure in a 10.0 liter flask at 27°C of a sample of gas that contains 6.0 grams of hydrogen, 15.2 grams of

motikmotik

Answer:

The total pressure is 27.8 atm

Explanation:

From the ideal gas equation,

PV = nRT

P (total pressure) = nRT/V

n (total moles of gases) = (6/1 moles of hydrogen) + (15.2/14 moles of nitrogen) + (16.8/4 moles of helium) = 6+1.1+4.2 = 11.3 moles

R = 0.082057L.atm/gmol.K, T = 27°C = 27+273K = 300K, V = 10L

P = 11.3×0.082057×300/10 = 27.8 atm

7 0

3 years ago

Determine the molar mass of CuSO4 (the solute) in a 1.0M aqueous solution of CuSO4

inna [77]

Answer:

See explanation.

Explanation:

Hello,

In this case, we could have two possible solutions:

A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

$M_{CuSO_4}=m_{Cu}+m_{S}+4*m_{O}=63.546 g/mol+32.00g/mol+4*16.00g/mol\\\\M_{CuSO_4}=159.546g/mol$

That is the mass of copper (II) sulfate contained in 1 mol of substance.

B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

$M=\frac{n_{solute}}{V_{solution}}$

So you can solve for the moles of the solute:

$n_{solute}=M*V_{solution}$

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

$n_{solute}=1mol/L*1.5L=1.5mol$