1. ____ is the measure of the pull of gravity on an object.

Which of these factors will cause a solid solute to dissolve faster?

Nostrana [21]

I believe the answer you are looking for is the 4th one.

8 0

3 years ago

Ammonia, nh3, is a weak base with a kb = 1.8 x 10-5. in a 0.8m solution of ammonia, which has a higher concentration: ammonia (n

insens350 [35]

Answer is: concentration ammonia is higher than concentration of ammonium ion.
Chemical reaction of ammonia in water: NH₃ + H₂O → NH₄⁺ + OH⁻.
Kb(NH₃) = 1,8·10⁻⁵.
c₀(NH₃) = 0,8 mol/L.
c(NH₄⁺) = c(OH⁻) = x.
c(NH₃) = 0,8 mol/L - x.
Kb = c(NH₄⁺) · c(OH⁻) / c(NH₃).
0,000018 = x² / 0,8 mol/L - x.
solve quadratic equation: x = c(NH₄⁺) = 3,79·10⁻³ mol/L.

7 0

2 years ago

AgI + KNO3<br>What type of reaction is this?

emmasim [6.3K]

Answer:

double replacement

Explanation:

4 0

2 years ago

A 256 mL sample of HCl gas is in a flask where it exerts a force (pressure) of 67.5 mmHg. What is the pressure of the gas if it

Effectus [21]

Answer:

The pressure in the new flask would be $128\; \rm mmHg$ if the $\rm HCl$ here acts like an ideal gas.

Explanation:

Assume that the $\rm HCl$ sample here acts like an ideal gas. By Boyle's Law, the pressure $P$ of the gas should be inversely proportional to its volume $V$ .

For example, let the initial volume and pressure of the sample be $V_1$ and $P_1$ . The new volume $V_2$ and pressure $P_2$ of this sample shall satisfy the equation: $P_1 \cdot V_1 =P_2 \cdot V_2$ .

In this question,

The initial volume of the gas is $V_1= 256\; \rm mL$ .
The initial pressure of the gas is $P_1 = 67.5\; \rm mmHg$ .
The new volume of the gas is $V_2 = 135\; \rm mL$ .

The goal is to find the new pressure of this gas, $P_2$ .

Assume that this sample is indeed an ideal gas. Then the equation $P_1 \cdot V_1 =P_2 \cdot V_2$ should still hold. Rearrange the equation to separate the unknown, $P_2$ . Note: make sure that the units for $V_1$ and $V_2$ are the same before evaluating. That way, the unit of

$\begin{aligned} & P_2\\ &= \frac{P_1 \cdot V_1}{V_2} \\ &= \frac{256\; \rm mL \times 67.5\; \rm mmHg}{135\; \rm mL} \\ & \approx 128\; \rm mmHg\end{aligned}$ .

3 0

3 years ago

238,2U --> 234Th +<br> 90

stiv31 [10]

Answer:

$\\_{2}^{4} \alpha$

Explanation:

In this question, we wish to find the missing nuclei for the equation:

$\\_{92}^{238} U\rightarrow _{90}^{234} Th + _{Z}^{A} X$

In order to find the missing species, we need to use the charge and mass balance law. That is, the mass should be conserved: the total mass on the left-hand side with respect to the arrow should be equal to the total mass on the right-hand side with respect to the arrow:

$238 = 234 + A$