How are electric fields different from gravitational fields?

Julli [10]

Energy which a body possesses by virtue of being in motion.

6 0

3 years ago

The value of AH° for the reaction below is +128.1 kJ CH3OH (I)-CO (g) + 2H2 (g) How much heat is consumed when 87.1 g of hydroge

natka813 [3]

Answer:

A) 2.76 x 103 kJ

Explanation:

CH3OH (I)--------->CO (g) + 2H2 (g)

Number of moles contained in 87.1g of hydrogen gas= mass of hydrogen gas/ molar mass of hydrogen gas

Molar mass of hydrogen gas= 2gmol-1

Number of moles hydrogen gas = 87.1g/2gmol-1= 43.55 moles of hydrogen

1 mole of methanol yields 2 moles of hydrogen

x moles of methanol yields 43.55 moles of hydrogen

x= 43.55 moles of hydrogen × 1 mole of methanol/2 moles of hydrogen

x= 21.775 moles of methanol

Then;

If 1 mole of methanol absorbs 128KJ of energy

21.775 moles of methanol will absorb 21.775 × 128/1 = 2.7×10^3 KJ of heat

6 0

3 years ago

How do you make a animal cell?

GenaCL600 [577]

Answer:you cant make an animal cell

Explanation:

8 0

3 years ago

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A cylindrical specimen of some metal alloy having an elastic modulus of 126 GPa and an original cross-sectional diameter of 3.4

Delvig [45]

Answer:

The maximum length of the specimen before deformation is 240.64 mm

Explanation:

Strain = stress ÷ elastic modulus

stress = load ÷ area

load = 2130 N

diameter = 3.4 mm = 3.4×10^-3 m

area = πd^2/4 = 3.142 × (3.4×10^-3)^2/4 = 9.08038×10^-6 m^2

stress = 2130 N ÷ 9.08038×10^-6 m^2 = 2.35×10^8 N/m^2

elastic modulus = 126 GPa = 126×10^9 Pa

Strain = 2.35×10^8 ÷ 126×10^9 = 0.00187

Length = extension ÷ strain = 0.45 mm ÷ 0.00187 = 240.64 mm

5 0

3 years ago

Calculate the ph of the resulting solution if 29.0 ml of 0.290 m hcl(aq) is added to

Mama L [17]

We have the given reaction as;

$HCl_{(aq)} + NaOH_{(aq)}$ ----> $NaCl_{(aq)} + H_{2}O_{(l)}$ 

Answer A) The pH will be 12.36,

We have to convert the concentrations of HCl and NaOH into moles,

So we have, n(HCl) = (0.0290 L) X (0.290 mol/L) = 8.41X $10^{-3}$ moles 

and for NaOH we have, n(NaOH) = (0.0390 L)(0.290 mol/L) = 1.13X $10^{-2}$ moles

Now, it seems NaOH is in excess, so amount remaining will be;

1.13 X $10^{-2}$ - 8.41 X $10^{-3}$ = 2.89 X $10^{-3}$ moles

Now, the total volume will become as = 0.0390 + 0.0290 = 0.068 L 

So, the concentration of [ $OH^{-}$ ] = 2.89×10ˉ³ mol / 0.068 L = 4.25 X $10^{-2}$ M

pOH = - log [ $OH^{-}$ ] = -log (4.25× $10^{-2}$ ) = 1.37

Hence, pH = 14 - pOH = 14 - 1.37 = 12.6

So the pH of the solution will be 12.6 which is basic in nature. 

Answer B) The pH will be 1.68

Now, for the given concentration we need to find moles for HCl and NaOH also;

n(HCl) = (0.0290 L)(0.290 mol/L) = 8.41 X $10^{-3}$ mol 

n(NaOH) = (0.0190 L)(0.390 mol/L) = 7.41 X $10^{-3}$ mol 

here we can see, HCl is in excess amount so the remaining will be;

8.41X $10^{-3}$ - 7.41 X $10^{-3}$ = 1.0 X $10^{-3}$ mol

Here, the total volume will be = 0.0290 + 0.0190 = 0.0480 L

So the concentration of [HCl] = 1.0 X $10^{-3}$ mol / 0.0480 L = 2.08 X $10^{-2}$ M

Which is = [H⁺] 

So, the pH = - log [ $H^{+}$ ] = -log(2.08X $10^{-2}$ ) = 1.68

Hence, the pH will be 1.63 which is more acidic in nature.

6 0

3 years ago

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