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3 years ago

What is the unit of measurement for the amount of electrical energy consumed during work?

Chemistry

1 answer:

Alinara [238K]3 years ago

4 0

Answer:

Joule

Electrical Energy: The Volt

Explanation:

As we now know that energy is the capacity to do work, with the standard unit used for energy (and work) being the Joule. A joule of energy is defined as the energy expended by one ampere at one volt, moving in one second.

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7. A figure skater skating across ice, who grabs another skater and brings him along the ice with her, is an example of what typ

xenn [34]

Answer:

The type of collision is A. Inelastic collision.

Explanation:

The macroscopic collisions are generally inelastic and do not conserve the kinetic energy, although of course the total energy is conserved. The inelastic collision is one in which the objects that collide remain together after the collision.

So, a figure skater skating across ice, who grabs another skater and brings him along the ice with her is a clear example of inelastic collision.

6 0

3 years ago

Read 2 more answers

_____ C8H8 + _____ O2 _____ CO2 + _____ H2O

Effectus [21]

What do you want for the answer the balance equation?

5 0

3 years ago

A materials scientist has created an alloy containing aluminum, copper, and zinc, and wants to determine the percent composition

liubo4ka [24]

Answer:

0.8749 grams of hydrogen gas was formed from the reaction.

Explanation:

P = Pressure of hydrogen gad= 744 Torr = 0.98 atm

(1 atm = 760 Torr)

V = Volume of hydrogen gas= 11 L

n = number of moles of hydrogen gas= ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of vapor = 27.0 °C = 300.15 K

Putting values in above equation, we get:

Using an ideal gas equation:

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.98 atm\times 11 L}{0.0821 L atm/mol K\times 300.15 K}$

n = 0.4374 moles

Mass of 0.4374 moles of hydrogen gas:

0.4374 mol × 2 g/mol = 0.8749 g

0.8749 grams of hydrogen gas was formed from the reaction.

7 0

3 years ago

When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H

algol [13]

Answer: The empirical formula is $CH_2$ and molecular formula is $C_4H_8$

Explanation:

We are given:

Mass of $CO_2$ = 18.95 g

Mass of $H_2O$ = 7.759 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 18.59 g of carbon dioxide, = $\frac{12}{44}\times 18.59=5.07g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 7.759 g of water, = $\frac{2}{18}\times 7.759=0.862g$ of hydrogen will be contained.

Mass of C = 5.07 g

Mass of H = 0.862 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.422}{0.422}=1$

For H = $\frac{0.862}{0.422}=2$

The ratio of C : H = 1: 2

Hence the empirical formula is $CH_2$ .

The empirical weight of $CH_2$ = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4$

The molecular formula will be= $4\times CH_2=C_4H_8$

6 0

3 years ago

Find [H+] of a 0.056 M hydrofluoric acid solution. Ka = 1.45 x 10-7

brilliants [131]

Answer: $[H^+]$ of 0.056 M HF solution is $8.96\times 10^{-5}$

Explanation:

$HF\rightarrow H^+F^-$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.056 M and $\alpha$ = ?

$K_a=1.45\times 10^{-7}$

Putting in the values we get:

$1.45\times 10^{-7}=\frac{(0.056\times \alpha)^2}{(0.056-0.056\times \alpha)}$

$(\alpha)=0.0016$

$[H^+]=c\times \alpha$

$[H^+]=0.056\times 0.0016=8.96\times 10^{-5}$

Thus $[H^+]$ of 0.056 M HF solution is $8.96\times 10^{-5}$

8 0

3 years ago