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2 years ago

1500 torr is how many kPa

Chemistry

2 answers:

Fiesta28 [93]2 years ago

5 0

Answer:

200 kPa

(it is actually 199.999, but rounding up it is 200)

hope this helps!

valkas [14]2 years ago

4 0

Answer:

1500 Torr is about 200 kPa

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6. List the different groups on the periodic table

Alex73 [517]

Answer:

Group 1: alkali metals, or lithium family.

Group 2: alkaline earth metals, or beryllium family.

Group 3: the scandium family.

Group 4: the titanium family.

Group 5: the vanadium family.

Group 6: the chromium family.

Explanation:

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3 years ago

A weather balloon at Earth’s surface has a

tigry1 [53]

The temperature : 263.016 K

<h3>Further explanation</h3>

Combined with Boyle's law and Gay Lussac's law

$\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}$

P1 = initial gas pressure (N/m² or Pa)

V1 = initial gas volume (m³)

P2 = gas end pressure

V2 = the final volume of gas

T1 = initial gas temperature (K)

T2 = gas end temperature

P1=760 mmHg

V1= 4 L

T1 = 275 K

P2=704 mmHg

V1=4.13 L

$\tt \dfrac{760\times 4}{275}=\dfrac{704\times 4.13}{T_2}\\\\T_2=263.016~K$

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3 years ago

Mr.Davies has requested that you wait for 10 more minutes is that informal or formal

LUCKY_DIMON [66]

Answer:

Formal

Explanation:

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4 years ago

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DENIUS [597]

The first whistle is louder while the second whistle has a higher pitch
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2 years ago

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Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and

rusak2 [61]

Answer:

$4Zn_(_s_)~+~7OH^-~_(_a_q_)~+~NO_3^-_(_a_q_)~+~6H_2O_(_l_)~-->4Zn(OH)_4^-^2_(_a_q_)~+~NH_3_(_g_)$

-) Oxidizing agent: $NO_3^-_(_a_q_)$

-) Reducing agent: $Zn_(_s_)$

Explanation:

The first step is separate the reaction into the semireactions:

A. $Zn~->Zn(OH)_4^-^2$

B. $NO_3^-~->~NH_3$

If we want to balance in basic medium we have to follow the rules:

1. We adjust the oxygen with $OH^-$

2. We adjust the H with $H_2O$

3. We adjust the charge with $e^-$

Lets balance the first semireaction A. :

$Zn~+~4OH^-~->Zn(OH)_4^-^2~+~2e^-$

Now, lets balance semireaction B:

$NO_3^-~+~8e^-~+~6H_2O~->~NH_3~+~9OH^-$

Finally, we have to add the two semireactions:

_________________________________________

$8~(Zn~+~4OH^-~->Zn(OH)_4^-^2~+~2e^-)$

$2~(NO_3^-~+~8e^-~+~6H_2O~->~NH_3~+~9OH^-)$

_________________________________________

$(8Zn~+~32OH^-~->8Zn(OH)_4^-^2~+~16e^-)$

$(2NO_3^-~+~16e^-~+~12H_2O~->~2NH_3~+~18OH^-)$

Cancel out the species on both sides:

$8Zn~+~14OH^-~+~2NO_3^-~+~12H_2O~-->8Zn(OH)_4^-^2~+~2NH_3$

Simplifying the equation :

$4Zn~+~7OH^-~+~NO_3^-~+~6H_2O~-->4Zn(OH)_4^-^2~+~NH_3$

The $Zn_(_s_)$ is oxidized therefefore is the reducing agent. The $NO_3^-_(_a_q_)$ is reduced therefore is the oxidizing agent.

4 0

4 years ago