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Vilka [71]
3 years ago
6

Chất tinh khiết là:

Chemistry
1 answer:
coldgirl [10]3 years ago
7 0

Answer:

The FitnessGram™ Pacer Test is a multistage aerobic capacity test that progressively gets more difficult as it continues. The 20 meter pacer test will begin in 30 seconds. Line up at the start. The running speed starts slowly, but gets faster each minute after you hear this signal. [beep] A single lap should be completed each time you hear this sound. [ding] Remember to run in a straight line, and run as long as possible. The second time you fail to complete a lap before the sound, your test is over.

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D water expands when it freezes

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What would ne the acceleration in a body moving wit uniform velocity and why​
meriva

Answer: The derivative of a constant term is always 0. So the acceleration of the body would be zero. Hence, the acceleration of a body moving with uniform velocity will always be zero.

Hope this helps :) :)

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Help me answer this question pls
Nostrana [21]
The answer should be D all of the above
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Un anillo, de masa 90 gramos, contiene 59,1% de oro. ¿Cuál es el valor del anillo si cada mol de oro vale S/.1800?. Considere el
LekaFEV [45]

Answer:

S/.486 es el valor del anillo

Explanation:

Para hallar el precio del anillo se deben encontrar las moles de oro que contiene este.

Si el anillo es de 90g y solo el 59.1% contiene oro, la cantidad de oro en gramos es:

90g × 59.1% = 53.19g Oro en el anillo

Ahora, para convertir los gramos de oro a moles se debe usar la masa atómica del oro (197g/mol), así:

53.19g × (1mol / 197g) = <em><u>0.27 moles de oro contiene el anillo</u></em>.

Ya que cada mol de oro cuesta S/.1800, 0.27 moles de oro (Y por lo tanto, el anillo) costarán:

0.27mol × (S/.1800 / 1mol oro) =

<h3>S/.486 es el valor del anillo</h3>
8 0
3 years ago
In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for
inessss [21]

Answer:

The standard cell potential of the reaction is 0.78 Volts.

Explanation:

Cu^{2+}(aq)+Fe(s)\rightarrow Cu(s)+Fe^{2+}(aq)

Reduction at cathode :

Cu^+(aq)+2e^-\rightarrow Cu(s)

Reduction potential of  Cu^{2+} to Cu=E^o_{1}=0.34 V

Oxidation at anode:

Fe(s)\rightarrow Fe^{2+}(aq)+2e^-

Reduction potential of  Fe^{2+} to Fe=E^o_{2}=-0.44 V

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}

Putting values in above equation, we get:

E^o_{cell}=0.34V -(-0.44 V)=0.78 V

The standard cell potential of the reaction is 0.78 Volts.

3 0
3 years ago
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