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3 years ago

What common household substances can turn blue litmus paper red?

1 answer:

5 0

There is a few things. The things that turn it red is acids.

Properties of acids include:
*Tasting sour
*Conduct electricity
*Corrosive

A few household acids would be lemons, vinegar, salt, and some sodas.

You might be interested in

What is the base number for the metric system of units? In other words, by what number do we multiply to move up in scale ? ____

Andru [333]

Answer:

Explanation:

Each time you want to move up the scale, the number is multiplied by 10. The opposite is true when moving down; you must divide by 10.

6 0

3 years ago

Read 2 more answers

Based on their locations on the periodic table, which best compares the properties of the metalloids arsenic (As) and antimony (

Len [333]

Arsenic has a more nonmetallic behavior than antimony because arsenic's atoms have fewer electron shells.

6 0

3 years ago

Read 2 more answers

A gas contained within a piston-cylinder assembly, initially at a volume of 0.1 m3 , undergoes a constant-pressure expansion at

Gnom [1K]

Answer:

Work: 4.0 kJ, heat: 4.25 kJ

Explanation:

For a gas transformation at constant pressure, the work done by the gas is given by

$W=p(V_f -V_i)$

where in this case we have:

$p = 2 bar = 2\cdot 10^5 Pa$ is the pressure

$V_i = 0.1 m^3$ is the initial volume

$V_f = 0.12 m^3$ is the final volume

Substituting,

$W=(2\cdot 10^5)(0.12-0.10)=4000 J = 4.0 kJ$

The 1st law of thermodynamics also states that

$\Delta U = Q-W$

where

$\Delta U$ is the change in internal energy of the gas

Q is the heat absorbed by the gas

Here we know that

$\Delta U = +0.25 kJ$

Therefore we can re-arrange the equation to find the heat absorbed by the gas:

$Q=\Delta U + W = 0.25 kJ + 4.0 kJ = 4.25 kJ$

7 0

4 years ago

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

3 years ago

A 14.0 gauge copper wire of diameter 1.628 mmmm carries a current of 12.0 mAmA . Part A What is the potential difference across

NARA [144]

Answer:

a) 2.063*10^-4

b) 1.75*10^-4

Explanation:

Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:

$A=\frac{\pi }{4}d^{2} \\=\frac{\pi }{4}*(1.628*10^-3 m)^2\\=2.082*10^-6 m^2\\$

a) The Potential difference across a 2.00 in length of a 14-gauge copper

wire:

L= 2.00 m

From Table Copper Resistivity $p$ = 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:

$R=\frac{pL}{A}$

=0.0165Ω

The Potential difference across the copper wire is:

V=IR

=2.063*10^-4

b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m

The Resistance of the Silver wire is:

$R=\frac{pL}{A}$

=0.014Ω

The Potential difference across the Silver wire is:

V=IR

=1.75*10^-4

4 0

3 years ago