1. A gas has a volume of 50.0 mL at a temperature of -73°C. What volume

If you have a voltage of 120 volts entering your house and a light bulb with 0.625 amperes flowing through it, how many watts is

Luba_88 [7]

Answer:

75 watts

Explanation:

4 0

3 years ago

Read 2 more answers

When the following reaction comes to equilibrium, will the concentrations of the reactants or products be greater? Does the answ

Nutka1998 [239]

Answer:

At equilibrium, the concentration of the reactants will be greater than the concentration of the products. This does not depend on the initial concentrations of the reactants and products.

Explanation:

The value of Kc gives us an idea of the extent of the reaction. A big Kc (Kc > 1) means that in the equilibrium there are more products than reactants, and the opposite happens for a small Kc (Kc < 1). The equilibrium is reached no matter what the initial concentrations are.

The value of the equilibrium constant is relatively SMALL; therefore, the concentration of reactants will be GREATER THAN the concentration of products. This result is INDEPENDENT OF the initial concentration of the reactants and products.

4 0

3 years ago

Write the balanced NET IONIC equation for the reaction that occurs when perchloric acid and calcium hypochlorite are combined. I

Elanso [62]

Answer:

$H^+(aq)+(ClO)^-(aq)\rightarrow HClO{(aq)}$

Explanation:

Hello there!

In this case, since perchloric acid is HClO4 and is a strong acid and calcium hypochlorite is Ca(ClO)2, the undergoing molecular chemical reaction turns out:

$2HClO_4{(aq)}+Ca(ClO)_2{(aq)}\rightarrow 2HClO{(aq)}+Ca(ClO_4)_2{(aq)}$

Thus, since the resulting hypochlorous acid is weak, it does not fully ionize, so it remains unionized, however, we can write the ions for the other species:

$2H^++2(ClO_4)^-+Ca^{2+}+2(ClO)^-\rightarrow 2HClO{(aq)}+Ca^{2+}+2(ClO_4)^-$

Now, we can cancel out the spectator ions, calcium and perchlorate, to obtain:

$2H^+(aq)+2(ClO)^-(aq)\rightarrow 2HClO{(aq)}\\\\H^+(aq)+(ClO)^-(aq)\rightarrow HClO{(aq)}$

Best regards!

3 0

3 years ago

Hello, everyone!

marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x $\frac{1 g}{1000 mg}$ x $\frac{1 mol}{28.01 g}$

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V = $\frac{0.00125 mol x 0.082057 \frac{atm L}{mol K} x 243 K}{0.92 atm}$

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x $\frac{1000 cm^{3} }{1 L}$ = 27.09 cm³

Then the acceptable concentration c of CO in ppm is,

c = 27 cm³ / m³ = 27 ppm

To express this concentration in percent by volume we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 $\frac{cm^{3} }{m^{3} }$ x $\frac{1 m^{3} }{1 000 000 cm^{3} }$ x 100%

c = 0.003 %

So, the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is 27.09 ppm and 0.003 %.

5 0

3 years ago

What will happen to the chemical equilibrium if mgci2 is added

gogolik [260]

Answer:

The chemical equilibrium of the system will be unaffected. The chemical equilibrium of the system will shift to the right to favor the forward reaction. The chemical equilibrium of the system will shift to the left to favor the reverse reaction. (I hope this helped!!)

7 0

3 years ago