I don’t know what this is this was a mistake?

The planet Mercury takes 176 days to do one spin around its pole. Compared to the Earth,

Tanya [424]

Answer:

The sun would appear to move more slowly across Mercury's sky.

Explanation:

This is because, the time it takes to do one spin or revolution on Mercury is 176 days (which is its period), whereas, the time it takes to do one spin or revolution on the Earth is 1 day.

Since the angular speed ω = 2π/T where T = period

So on Mercury, T' = 176days = 176 days × 24 hr/day × 60 min/hr × 60 s/min = ‭15,206,400‬ s

So, ω' = 2π/T'

= 2π/15,206,400‬ s

= 4.132 × 10⁻⁷ rad/s

So on Earth, T" = 1 day = 1 day × 24 hr/day × 60 min/hr × 60 s/min = ‭‭86,400‬ s

So, ω" = 2π/T"

= 2π/86,400‬ s

= 7.272 × 10⁻⁵ rad/s

Since ω' = 4.132 × 10⁻⁷ rad/s << ω" = 7.272 × 10⁻⁵ rad/s, <u>the sun would appear to move more slowly across Mercury's sky.</u>

4 0

3 years ago

what is the molecular formula of a compound with an empiricla formula of ch and a molecular mass of 78

Dovator [93]

Answer:

C6H6

Explanation:

We can obtain the molecular formula from the empirical formula.

What we need do here is:

(CH)n = 78

The n shows the multiples of both element present in the actual compound. It can be seen that carbon and hydrogen have the same element ratio here. We then use the atomic masses of both elements to get the value of n. The atomic mass of carbon is 12 a.m.u while the atomic mass of hydrogen is 1 a.m.u

(1 + 12)n = 78

13n = 78

n = 78/13 = 6

The molecular formula is

(CH)n = (CH)6 = C6H6

6 0

3 years ago

Se sabe que 10 g de calcio reaccionan con 4 g de oxígeno para obtener 14 g de óxido de calcio. Indica la cantidad de óxido de ca

egoroff_w [7]

Answer:

Si se usan 50 gramos de calcio y óxigeno, se obtienen 70 gramos de óxido de calcio.

Explanation:

Hola,

En este caso, la reacción llevada a cabo es:

$2Ca+O_2\rightarrow 2CaO$

De este modo si asumimos el ejemplo dado, 50 gramos de calcio, cuya masa atómica es 40 g/mol y 50 g de oxígeno, cuya masa atómica como gas diatómico es 32 g/mol, antes de calcular los gramos de óxido de calcio producidos, debemos identificar el reactivo límite. Así, calculamos las moles de calcio disponibles en 50 g:

$mol_{Ca}^{disponible}=50gCa*\frac{1molCa}{40gCa} =1.25molCa$

Y también las moles de calcio consumidas por los 50 g de oxígeno, utilizando su relación molar 2:1:

$mol_{Ca}^{consumidas\ por\ O_2}=50gO_2*\frac{1molO_2}{32gO_2} *\frac{2molCa}{1molO_2} =3.125molCa$

Por lo tanto, hay menos calcio disponible que el que consume el oxígeno, por lo que el calcio esel reactivo límite. Ahora, con este, calculamos los gramos de óxido de calcio, cuya masa molar es 56 g/mol, que se producen:

$m_{CaO}=1.25molCa*\frac{2molCaO}{2molCa}* \frac{56gCaO}{1molCaO}\\ \\m_{CaO}=70gCaO$