How many lone pairs are present in the molecule SeBr2? Can someone explain please?

An increase in greenhouse gases results in?

grin007 [14]

An increase in the atmospheric concentrations of greenhouse gases produces a positive climate forcing, or warming effect. From 1990 to 2015, the total warming effect from greenhouse gases added by humans to the Earth's atmosphere increased by 37 percent.

5 0

3 years ago

If i initially have a gas with a pressure of 84 kpa and a temperature of 350 c and i heat it an additional 230 degrees, what wil

Ganezh [65]

Answer:

67.824

Explanation: You want to use the combined gas law equation (P1*V1)/(n1*T1)=(P2*V2)/(n2*T2). So first cross out what remains constant, so volume(V) and I assume moles (since it was not mentioned as a change). Then you can solve algebraically for the answer!

Hope this helped!

5 0

3 years ago

Considering the following precipitation reaction: Pb(NO3)2(aq) + 2Kl(aq) – Pbl (s) + 2KNO3(aq) IN What is the correct complete i

brilliants [131]

Answer:

Pb²++2I−→PbI2(s)

yellow precipitate

Explanation:

6 0

3 years ago

When the metallic element sodium combines with the nonmetallic element bromine, br2(l), what is the chemical formula of the prod

inysia [295]

The element Sodium will form a +1 cation, while the bromine a -1 anion. Since the anion bromine is always a liquid, therefore its symbol is Br2(l). So, the chemical equation for their reaction is: 
2Na(s) + Br2(l) -> 2NaBr(aq)

5 0

3 years ago

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Some SO2 and O2 are mixed together in a flask at 1100 K in such a way ,that at the instant of mixing, their partial pressures ar

kakasveta [241]

Answer:

The answer is " $\bold{0.525\ \ atm^{-1}}$ "

Explanation:

Given equation:

$2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g)$

Given value:

$\Delta rH =-198.2 \ \ \frac{KJ}{mol}$

$Kp=1100 \ K$

$\Delta x = 2-(2+1)\\\\$

$= 2-(2+1)\\\\= 2-(3)\\\\= -1$

$\left\begin{array}{cccc}I\ (atm)&1&0.5&0\\C\ (atm)&2x&-x&2x\\E\ (atm) &1-2x&0.5-x&2x\end{array}\right$

calculating the total pressure on equilibrium= $(1-2x)+(0.5-x)+2x \ atm\\\\$

$= 1-2x+0.5-x+2x\\\\= 1+0.5-x\\\\=1.5-x\\$

$\therefore \\\\\to 1.50-x= 1.35 \\\\\to 1.50-1.35= x\\\\\to 0.15= x\\\\ \to x= 0.15\\\\$

calculating the pressure in $So_2$ :

$= (1-2 \times 0.15)$

$= 1-0.30 \\\\ =0.70 \ atm$

calculating the pressure in $O_2$ :

$= (0.5- 0.15)\\\\= 0.35 \ atm \\$

calculating the pressure in $So_3$ :

$= (2 \times 0.15)\\\\= (.30) \ atm \\\\$

Calculating the Kp at 1100 K:

$= \frac{(Pressure(So_3))^2}{(Pressure(So_2))^2 \times Pressure(O_2)}\\\\= \frac{0.30^2}{0.70^2 \times 0.35}\\\\= \frac{0.30 \times 0.30 }{0.70\times 0.70 \times 0.35}\\\\= \frac{0.09 }{0.49\times 0.35} \\\\= \frac{0.09 }{0.1715} \\\\= 0.5247 \ \ or \ \ 0.525 \ \ atm^{-1} \\\\$

4 0

4 years ago