Please help its super hard

The volume of a gas is 0.30 L when the pressure is 4.00 atm. At the same temperature,

padilas [110]

Answer:

The presion is 0.6 atm

Explanation:

P1V1=P2V2

P2 = P1V1/V2

P2 = (4.00 atm * 0.30 L) / 2.0 L

P2= 0.6 atm

7 0

4 years ago

What's the highest electronegativity

TiliK225 [7]

Your answer is H) florine

3 0

3 years ago

Read 2 more answers

What’s the balanced equation for the reaction between zinc oxide and dilute hydrochloric acid ?

larisa [96]

Answer:

ZnO + 2HCl → ZnCl₂ + H₂O

Explanation:

The unbalanced reaction between zinc oxide (ZnO) and dilute hydrochloric acid (HCl) is:

ZnO + HCl → ZnCl₂ + H₂O

Now we'll procede to balance the equation:

There is 1 Cl atom on the left side and 2 on the right, so we change that:

ZnO + 2HCl → ZnCl₂ + H₂O

The number of atoms for each element is the same on both sides, so the equation is now balanced.

7 0

3 years ago

Coulombic attraction is the attraction between what two particles in the atom?

tankabanditka [31]

Answer:

እርስዎ ፊት ላላ ውስጥ ጡት ነዎት

እርስዎ ፊት ላላ ውስጥ ጡትExplanation:

4 0

3 years ago

What is the equilibrium partial pressure of water vapor above a mixture of 62.9 g H2O and 33.2 g HOCH2CH2OH at 55 °C. The partia

bija089 [108]

Answer : The partial pressure of $H_2O$ is 102.3 mmHg.

Explanation :

As per question,

Mass of $H_2O$ = 62.9 g

Mass of $HOCH_2CH_2OH$ = 33.2 g

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $HOCH_2CH_2OH$ = 62 g/mole

First we have to calculate the moles of $H_2O$ and $HOCH_2CH_2OH$ .

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{62.9g}{18g/mole}=3.49mole$

$\text{Moles of }HOCH_2CH_2OH=\frac{\text{Mass of }HOCH_2CH_2OH}{\text{Molar mass of }HOCH_2CH_2OH}=\frac{33.2g}{62g/mole}=0.536mole$

Now we have to calculate the mole fraction of $H_2O$ and $HOCH_2CH_2OH$ .

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }HOCH_2CH_2OH}=\frac{3.49}{3.49+0.536}=0.867$

$\text{Mole fraction of }HOCH_2CH_2OH=\frac{\text{Moles of }HOCH_2CH_2OH}{\text{Moles of }H_2O+\text{Moles of }HOCH_2CH_2OH}=\frac{0.536}{3.49+0.536}=0.134$

Now we have to partial pressure of $H_2O$ .

According to the Raoult's law,

$p^o=X\times p_T$

where,

$p^o$ = partial pressure of water vapor

$p_T$ = total pressure of gas

$X$ = mole fraction of water vapor

$p_{H_2O}=X_{H_2O}\times p_T$

$p_{H_2O}=0.867\times 118.0mmHg=102.3mmHg$

Therefore, the partial pressure of $H_2O$ is 102.3 mmHg.

3 0

3 years ago