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natima [27]
3 years ago
10

The SI unit of length equal to the distance light travels in a vacuum in 1/299 792 458 of a second is the ___.

Chemistry
1 answer:
ella [17]3 years ago
3 0

Answer:

Meter.

Explanation:

Hello,

In this case, The SI unit of length equal to the distance light travels in a vacuum in 1/299 792 458 of a second is the meter as the physical standard used to establish such length.

Regards.

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Which factor can increase the potential energy of an object?
OLEGan [10]

Answer:

Increase in height from the ground.

Explanation:

Potential energy =mass×acceleration due to gravity×height.

P.E= mgh

4 0
3 years ago
Which explains the information needed to calculate speed and velocity?
USPshnik [31]

Answer:

Both require time, but velocity requires displacement and speed requires distance

Explanation:

For calculating speed we require time and distance because speed is defined as the distance per unit time and as speed is a scalar quantity it does not have any direction

But for calculating the velocity we require time as well as displacement because velocity is defined as the displacement per unit time and as velocity is a vector quantity it has direction

Displacement is the shortest distance between the initial position and the final position and it has a specified direction as well

8 0
3 years ago
Read 2 more answers
Calculate the unit cell edge length for an 78 wt% Ag- 22 wt% Pd alloy. All of the palladium is in solid solution, and the crysta
BaLLatris [955]

Answer:

The unit cell edge length for the alloy is 0.405 nm

Explanation:

Given;

concentration of Ag, C_{Ag} = 78 wt%

concentration of Pd, C_P_d = 22 wt%

density of Ag = 10.49 g/cm³

density of Pd = 12.02 g/cm³

atomic weight of Ag, A_A_g = 107.87 g/mol

atomic weight of iron, A_P_d = 106.4 g/mol

Step 1: determine the average density of the alloy

\rho _{Avg.} = \frac{100}{\frac{C_A_g}{\rho _A_g} + \frac{C__{Pd}}{\rho _{Pd}} }

\rho _{Avg.} = \frac{100}{\frac{78}{10.49} + \frac{22}{12.02} } = 10.79 \ g/cm^3

Step 2: determine the average atomic weight of the alloy

A _{Avg.} = \frac{100}{\frac{C_v}{A _A_g} + \frac{C__{Pd}}{A _{Pd}} }

A _{Avg.} = \frac{100}{\frac{78}{107.87} + \frac{22}{106.4} } = 107.54 \ g/mole

Step 3: determine unit cell volume

V_c=\frac{nA_{avg.}}{\rho _{avg.} N_a}

for a FCC crystal structure, there are 4 atoms per unit cell; n = 4

V_c=\frac{4*107.54}{ 10.79*6.022*10^{23}} = 6.620*10^{-23} \ cm^3/cell

Step 4: determine the unit cell edge length

Vc = a³

a = V_c{^\frac{1}{3} }\\\\a = (6.620*10^{-23}}){^\frac{1}{3}}\\\\a = 4.05 *10^{-8} \ cm= 0.405 nm

Therefore, the unit cell edge length for the alloy is 0.405 nm

7 0
3 years ago
HELP !<br> 85g of NiCl2 dissolves in 350ml of H2O. What is the Molarity? Show your work
trapecia [35]

Answer:

1.874 M.

Explanation:

<em>Molarity is defined as the no. of moles of a solute per 1.0 L of the solution.</em>

<em />

M = (no. of moles of solute)/(V of the solution (L)).

<em>∴ M = (mass/molar mass)of NiCl₂/(V of the solution (L)).</em>

<em></em>

∴ M = (mass/molar mass)of NiCl₂/(V of the solution (L)) = (85.0 g / 129.59 g/mol)/(0.35 L) = 1.874 M.

7 0
3 years ago
Write the empirical formula of at least four binary ionic compounds that could be formed from the following ions:
cupoosta [38]
MgO
MgI2
PbO2
PbI4

These are the possible compounds
4 0
3 years ago
Read 2 more answers
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