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2 years ago

Which two technologies use reflected sound waves

Chemistry

1 answer:

katen-ka-za [31]2 years ago

4 0

Answer: The answer is B.Sonar and C.Radar

Explanation:

They are both just like an echo which is the main things that uses reflected sound waves. They are basically like synonyms for echo.

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A student heats a sample of hydrate once, and the mass of the sample and the evaporating dish is 16.428 g. After a second heatin

jolli1 [7]

Answer:

12.371 g

Explanation:

Given :

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g$

$m_{evaporating\ dish}+m_{First\ heated\ sample}=16.428\ g$

$m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g$

Mass of salt hydrate:

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g$

$m_{Hydrate\ sample}=25.637-m_{evaporating\ dish}\ g=25.637-1.135\ g=24.502\ g$

Mass of salt anhydrous:

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g$

$m_{Second\ heated\ sample}=m_{salt\ anhydrous}=13.266-m_{evaporating\ dish}\ g=13.266-1.135\ g=12.131\ g$

Mass of water:

$m_{water}=m_{Hydrate\ sample}-m_{salt\ anhydrous}=24.502-12.131\ g=12.371\ g$

$m_{water}=12.371\ g$

4 0

3 years ago

¿ ppr que no podemos automedicarnos ,cuando pensamos q tenemos el virus del CIVD-19?

GREYUIT [131]

Answer:

Porque como tu no eres profesional en medicina, no sabes nada sobre las dosis de medicamentos ni de los efectos que pueden tener en tu cuerpo si se administran incorrectamente.

Explanation:

Es logico que solo vayas al medico para que te den la cura allí.

3 0

3 years ago

The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3

9966 [12]

Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Where,

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol

$k_2=2.3\times 10^8$

$T_2=280\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (280 + 273.15) K = 553.15 K

$T_2=553.15\ K$

$k_1=4.8\times 10^8$

$T_1=376\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (376 + 273.15) K = 649.15 K

$T_1=649.15\ K$

So,

$\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)$