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4 years ago

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What is half-life? a. Half-life is a term that describes the time it takes for half of a particle to disintegrate c. Half-life i

s a term that describes the time it takes for the amount of radioactivity to increase by one half. b. Half-life is a term that describes the time it takes for the amount of radioactivity to go down by one half. d. None of the above

1 answer:

Alisiya [41]4 years ago

4 0

Well, half-life is the radioactivity of an identified isotope that decreases by half of the actual value.
So, your answer would be C.

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What relation does the boiling point of an amine have to a similar hydrocarbon?

den301095 [7]

Answer:

Amine have higher boiling points than hydrocarbons.

Explanation:

Primary, secondary and tertiary amines have higher boiling points than hydrocarbons because they can engage in intermolecular hydrogen bonding.

Amines has three classes

1. Primary amines

2. Secondary amines

3. Tertiary amines

All this classes of amines have higher boiling point than hydrocarbons due to C-N bond in them

This is because amines can engage in hydrogen bonding with water, amines of low molar mass are quite soluble in water.

Amines are having higher boiling points than hydrocarbons, as C-N bond in amines is more polar than a C-C bond in hydrocarbons. Due to the polar nature of amines, it forms intermolecular H-bonds and exists as associated molecules.

5 0

3 years ago

Read 2 more answers

9. Una jeringa contiene cloro gaseoso, que ocupa un volumen de 95 mL a una presión de 0,96 atm. ¿Qué presión debemos ejercer en

masha68 [24]

Answer:

2.61 atm

Ley de Boyle

Explanation:

$P_1$ = Presión inicial = 0.96 atm

$P_2$ = Presión final

$V_1$ = Volumen inicial = 95 mL

$V_2$ = Volumen final = 35 mL

En este problema usaremos la ley de Boyle.

$\dfrac{P_1}{P_2}=\dfrac{V_2}{V_1}\\\Rightarrow P_2=\dfrac{P_1V_1}{V_2}\\\Rightarrow P_2=\dfrac{0.96\times 95}{35}\\\Rightarrow P_1=2.61\ \text{atm}$

La presión ejercida sobre el émbolo para reducir su volumen es de 2.61 atm.

4 0

2 years ago

Latitude and longitude picture.

TiliK225 [7]

Answer:

where? please comment it

Explanation:

7 0

3 years ago

The law of inertia to both moving and no moving objects. True or false

Zepler [3.9K]

True, the law of inertia effects both moving and non-moving objects.

3 0

3 years ago

Read 2 more answers

A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What

AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, $\alpha=3.3 rad/s^2$

Diameter of the wheel, d=21 cm

Radius of wheel, $r=\frac{d}{2}=\frac{21}{2}$ cm

Radius of wheel, $r=\frac{21\times 10^{-2}}{2} m$

1m=100 cm

Magnitude of total linear acceleration, a= $1.7 m/s^2$

We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration, $a_t=\alpha r$

$a_t=3.3\times \frac{21\times 10^{-2}}{2}$

$a_t=34.65\times 10^{-2}m/s^2$

Radial acceleration, $a_r=\frac{v^2}{r}$

We know that

$a=\sqrt{a^2_t+a^2_r}$

Using the formula

$1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}$

Squaring on both sides

we get

$2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}$

$\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}$

$v^4=r^2\times 2.7699$

$v^4=(10.5\times 10^{-2})^2\times 2.7699$

$v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}$

$v=0.418 m/s$

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

6 0

3 years ago