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4 years ago

12

Sodium and Chlorine combine to form table salt. Given the chemical reaction, how many grams of 2Na + Cl2 would be required to pr

oduce 4 grams of table salt?

2 answers:

Darina [25.2K]4 years ago

5 0

The answer is 4 grams because matter can neither be created or destroyed and according to the laws of physics, you need 4 grams for 4 grams.

laila [671]4 years ago

4 0

4 GRAMS I had it on USATestPrep

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In the demolition of an old building, a 1,300 kg wrecking ball hits the building at 1.07 m/s2

goldfiish [28.3K]

Answer:

4

Explanation:

TRUST ME AND I AM SORRY IF RONG

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3 years ago

A merchant in Katmandu sells you a solid gold 1.00-kg statue for a very reasonable price. When you get home, you wonder whether

White raven [17]

Answer:

a) 51.8 cm³

b) kg/m³ is a dimension of density (mass/volume). The regular unitys for volume are m³, cm³, L, gallons.

Explanation:

a) The density of pure gold is 19.3 g/cm³. When put in water, the piece of gold will occupy a volume, so that the volume of water will be displaced. To know the volume, we must divide the mass for the density (mass must be in grams because of the units of the density)

V = 1000/19.3

V = 51.8 cm³

7 0

4 years ago

Newton’s law of cooling states that dx dt = −k(x−A) where x is the temperature,t is time, A is the ambient temperature, and k &g

lys-0071 [83]

Answer:

a) $X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}$

b)Does not affect the long term.

Explanation:

Given that

$\dfrac{dx}{dt}=-k(x-A)$

A = A0 cos(ωt)

$\dfrac{dx}{dt}=-k(x-A_o cos(\omega t))$

$\dfrac{dx}{dt}+kx=kA_o cos(\omega t)$

This is linear equation so integration factor ,I

$I=e^{\int kdt}$

$I=e^{kt}$

Now by using linear equation property

$e^{kt} X=\int e^{kt} kA_o cos(\omega t) dt +C$

$e^{kt} X= kA_o \dfrac{e^{kt}}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+C$

$X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}$

b)

at t= 0

$X(0)=\dfrac{k^2A_o}{\omega^2+k^2}+C$

$X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+e^{-kt}\times \left ( X(0)-\dfrac{k^2A_o}{\omega^2+k^2} \right )$

So the initial condition does not affect the long term.

5 0

4 years ago

Ohms law? explanation.

shusha [124]

Answer:

$\boxed{ \bold{ \sf{see \: below}}}$

Explanation:

$\boxed{ \bold{ \underline{ \huge{ \boxed{ \sf{ohm's \: law}}}}}}$

The relation between current through a metallic conductor and potential difference across its ends was studied systematically by a German physicist , <u>George </u><u>Simon </u><u>Ohm </u> in <u>1</u><u>8</u><u>2</u><u>6</u><u> </u><u>AD </u>. This relation is now known as Ohm's law. It states that <u>the </u><u>electric </u><u>current </u><u>passing </u><u>through</u><u> </u><u>a </u><u>conductor </u><u>is </u><u>directly </u><u>proportional</u><u> </u><u>to </u><u>the </u><u>potential </u><u>difference</u><u> </u><u>across </u><u>its </u><u>two </u><u>ends </u><u>at </u><u>a </u><u>constant </u><u>physical </u><u>condition</u><u> </u><u>[</u><u> </u><u>Temperature</u><u> </u><u>,</u><u> </u><u>cross </u><u>-</u><u> </u><u>sectional </u><u>area </u><u>,</u><u> </u><u>length </u><u>,</u><u> </u><u>nature </u><u>of </u><u>material </u><u>etc </u><u>]</u>

Hope I helped!

Best regards!!

4 0

4 years ago

4) A force of 500 N acts on an area of 0.05m2. Find the pressure in Pascal.

snow_tiger [21]

Answer:

pressure = force ie 500 N divided by area ie 0.05m².

p=f by a

p= 500n divided by 0.05 m²

p= 10,000 pascal

6 0

3 years ago