Help me with my bell work

Chemistry

2 answers:

Triss [41]3 years ago

6 0

1.) The four inner planets have slower orbits and the four outer planets have faster orbits.

2.) The closer a planet is to the Sun, the stronger the Sun's gravitational pull on it, and the faster the planet moves.

3.) Gravity

blagie [28]3 years ago

4 0

Answer:

1) It gets slower the farther out they are.

2) The farther out you are from the sun the less gravitational pull you have, which makes it go slower as it orbits.

3) Gravity ;) :)

Explanation:

Hope this helps! Plz mark as brainliest! :)

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Write Qc for each of the following:(1) Gaseous sulfur tetrafluoride reacts with liquid water to produce gaseous sulfur dioxide a

snow_tiger [21]

Answer : The expression for reaction quotient will be :

(1) $Q_c=\frac{[SO_2][HF]^4}{[SF_4]}$

(2) $Q_c=\frac{[O_2]^2[Xe]}{[XeF_2]}$

Explanation :

Reaction quotient $(Q_c)$ : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

(1) The given balanced chemical reaction is,

$SF_4(g)+2H_2O(l)\rightarrow SO_2(g)+4HF(g)$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted. So, the expression for reaction quotient will be :

$Q_c=\frac{[SO_2][HF]^4}{[SF_4]}$

(2) The given balanced chemical reaction is,

$2MoO_2(s)+XeF_2(g)\rightarrow 2MoF(l)+Xe(g)+2O_2(g)[/texIn this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted. So, the expression for reaction quotient will be :[tex]Q_c=\frac{[O_2]^2[Xe]}{[XeF_2]}$

7 0

3 years ago

A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f

a_sh-v [17]

Answer:

$m_{H_2O}=39.0g$

Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

$\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})$

Thus, both iron's and water's heat capacities are: 0.444 and 4.18 J/g°C respectively, so one solves for the mass of water as shown below:

$m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g$