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2 years ago

What was the main purpose of the opportunity rover?

Chemistry

2 answers:

pogonyaev2 years ago

8 0

Answer:

Explanation:The major goals of the two rovers, according to NASA, were to determine whether life as we know it could ever have arisen on Mars (focusing particularly on searching for ancient water) and characterizing the climate and geology of

Alex Ar [27]2 years ago

7 0

Answer:

The major goals of the two rovers, according to NASA, were to determine whether life as we know it could ever have arisen on Mars (focusing particularly on searching for ancient water) and characterizing the climate and geology of the planet

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A reaction between an acid and a base produced lithium chloride (LiCl). What acid and base combination could produce this salt?

erica [24]

Answer:

B. hydrochloric acid and lithium hydroxide

Explanation:

4 0

2 years ago

What determines how much energy an area receives

Lemur [1.5K]

The more focused the rays are, the more energy an area receives, and the warmer it is. The lowest latitudes get the most energy from the Sun. The highest latitudes get the least. The difference in solar energy received at different latitudes drives atmospheric circulation.

7 0

3 years ago

A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

1. Calculate the thermometer readings after 0.5 min and 1 min

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

t = t - 1, because the cooling starts 1 min late

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}$

3. Plot the temperature readings as a function of time.

The graphs are shown below.

6 0

3 years ago

Determine the value of the equilibrium constant, Kgoal, for the reactionN2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=?by making use of th

Black_prince [1.1K]

Answer:

$K_{goal}=1.793*10^{-33}$

Explanation:

N2(g)+O2(g)⇌2NO(g), $K_1 = 4.10*10^{-31}$

N2(g)+2H2(g)⇌N2H4(g), $K_2 = 7.40*10^{-26}$

2H2O(g)⇌2H2(g)+O2(g), $K_3 = 1.06*10^{-10}$

If we add above reaction we will get:

2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g) Eq (1)

Equilibrium constant for Eq (1) is $K_1*K_3*K_3$

Divide Eq (1) by 2, it will become:

N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g) Eq (2)

Equilibrium constant for Eq (2) is $(K_1*K_3*K_3)^{1/2}$

$Equilibrium constant =K_{goal}= (K_1*K_2*K_3)^{1/2}\\K_{goal}= (4.10*10^{-31} *7.40*10^{-26}*1.06*10^{-10})^{1/2}\\K_{goal}=1.793*10^{-33}$

7 0

3 years ago

What makes hydrogen peroxide a polar molecule?

kupik [55]

<span>C. polar bonds and asymmetrical structure
If the molecule contains polar bonds but it has a symmetrical structure, the polar bonds will cancel each other out so the overall molecule will be non-polar.
On the other hand, if the molecule contains polar bonds but has an asymmetrical structure, then the polar bonds won't cancel each other out, so the overall molecule ends up being polar.
</span>

7 0

3 years ago

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