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3 years ago

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Which actions are evidence that a chemical reaction has occurred? Select five options. a change in color the formation of a prec

ipitate a clear liquid becoming cloudy a change in shape the formation of bubbles the formation of smoke

1 answer:

vladimir1956 [14]3 years ago

8 0

Answer:

A change in color

The formation of a precipitate

A clear liquid becoming cloudy

The formation of bubbles

The formation of smoke

Explanation:

A chemical reaction can be defined as the conversion of reactant into product. The process in which one or more than one substance is converted into one or many different substances.

There are some indications which can provide the evidence that chemical reaction has occurred.

A change in color of the substance, formation or smoke or bubbles. The formation of the precipitate and in case the substance becomes cloudy.

All these evidences show that chemical reaction has occurred.

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27 degrees to kelvin, 20 dg to milligrams, and 3 to decimeters.

Gala2k [10]

K = C + 273, so 27°C = 27+273 = 300 K

1 dg = 100 mg, so 20 dg = 20×100 = 2,000 mg

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3 years ago

Calculate the force that the 4kg block exerts on the 10kg block

Kryger [21]

Acceleration=force/mass=28/(10+4)=2m/s^2

force10kg=ma=10*2
force4kg=ma=(10*2)=20
the4 kg is pushing against the 10kg block

vf=vi+at
-10=20*28/14 * t
t=30/2=15sec

i hope this can help you.

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Two objects, A and B, each of mass o.22 kg, are moving at 0.34 m/s directly toward each other. What is the speed of Object A aft

ki77a [65]

Theoretically, if the objects have the same mass and are moving towards each other at a speed of $0.34m s^{-1}$ , after a perfectly elastic collision, the object A is supposed to move with the same velocity in the opposite direction.

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3 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

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Which of these activities increases the amount of carbon in the atmosphere? A. animals eating plants B. burning of fossil fuels

tester [92]

Answer:

B. Burning of fossil fuels

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3 years ago