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3 years ago

Helppp nowwww plssssss!!

Chemistry

1 answer:

Anna007 [38]3 years ago

4 0

PLS HELP ME ASAP I DONT HAVE TIME. IT ALSO DETECTS IF ITs RIGHT OR WRONG. PLS HELP ME ASAP I DONT HAVE TIME. IT ALSO DETECTS IF ITs RIGHT OR WRONG. PLS HELP ME ASAP I DONT HAVE TIME. IT ALSO DETECTS IF ITs RIGHT OR WRONG. PLS HELP ME ASAP I DONT HAVE TIME. IT ALSO DETECTS IF ITs RIGHT OR WRONG. PLS HELP ME ASAP I DONT HAVE TIME. IT ALSO DETECTS IF ITs RIGHT OR WRONG. PLS HELP ME ASAP I DONT HAVE TIME. IT ALSO DETECTS IF ITs RIGHT OR WRONG. PLS HELP ME ASAP I DONT HAVE TIME. IT ALSO DETECTS IF ITs RIGHT OR WRONG. PLS HELP ME ASAP I DONT HAVE TIME. IT ALSO DETECTS IF ITs RIGHT OR WRONG.

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Water is heated to boiling and water vapor (steam) is released. Chemical change or physical change?

Vlad [161]

Physical because it is still H2O

6 0

3 years ago

Find the density.... Mass is 138g and Volume is 100 mL

solniwko [45]

Answer:

1380 kilogram/cubic meter

$p=\frac{m}{v}$

$=\frac{138g}{100mL}$

$=1.38$

$=1380$

4 0

3 years ago

When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

3 0

3 years ago

Which describes a way to speed up the collisions between hydrogen and oxygen molecules to produce more water?

rodikova [14]

Equation:2H2(g) + O2(g) → 2H2O(g)

Smaller container means less volume, and the molecules will hit the walls of the container more frequently because there's less space available and the pressure will go up. I guess this would mean that the side with fewer moles would be favored as a result. We count the number of moles on the reactants and products and find that there are fewer moles on the product side, so I guess this would favor the product formation.

4 0

3 years ago

Read 2 more answers

Consider this equilibrium reaction between carbon monoxide and hydrogen gas, occurring in a sealed flexible container. CO(g) + 3

wariber [46]

Answer:

More H2(g) is added to the container : Towards products.

CO is removed from the container : Towards reactants.

More CH4(g) is added to the container : Towards reactants

H2O(g) is removed from the container : Towards products.

The contents of the container are heated up. : Towards the reactants.

The contents of the container are cooled down : Towards the products.

The pressure inside the container is increased. :Towards the products

The container is stretched to increase the volume: Towards the reactants.

Explanation: :

CO(g) + 3 H2g) → CH4(g) + H2O(g)+ heat

There is released heat, so this reaction is exothermic

If the H2 concentration is increased, the system will try to change the concentration change by shifting the balance to the right, and thus the concentration of products will increase. Towards products.

If the CO is removed, the system will try to change this situation by shifting the balance to the left, and thus the concentration of reactants will increase, the concentration of products will decrease. Towards reactants.

If the CH4 concentration is increased, the system will try to change the concentration change by shifting the balance to the left, and thus the concentration of reactants will increase. Towards reactants

If the H2O is removed, the system will try to change this situation by shifting the balance to the right, and thus the concentration of products will increase, the concentration of products will decrease. Towards products.

If the temperature is increased, the system will reduce the amount of heat released. So the balance will shift to the left. Towards the reactants.

This because the extra heat / energy must be used.

If the temperature is decreased, the system will produce more heat So the balance will shift to the right. Towards the products.

This because more heat /energy needs to be produced to make up for the loss of heat (energy).

If the pressure is increased, the system will shift to the side with fewer moles of gas. In this case, there are 4 moles on the left and 2 moles on the right. So the balance will shift to the right. Towards the products. An increase of pressure has the same effect on the equilibrium as a decrease of the volume.

If the volume is increased, this means the pressure is decreased, the system will shift to the side with most moles of gas. In this case, there are 4 moles on the left and 2 moles on the right. So the balance will shift to the left. Towards the reactants. An increase of volume has the same effect on the equilibrium as a decrease of the pressure.

6 0

3 years ago