Two reactants, each having a mass of 50.0 g, react fully in a beaker. What is the total mass of the products in the reaction?

compare a humans changes during exercise to the changes that occur a during a dogs exercise when it is exposed to a hot environm

svetoff [14.1K]

A human body, just like a dog's, will sweat. Dogs will pants and sweat through the pads of their feet to cool down, and human will sweat through their foreheads, armpits, etc.
Dogs will tend to, in hot environments, lay on the floor or where the surface is cooler. Since they cannot simply strip their clothing to keep cool they tend to find cool surfaces, fans, sources of air, etc. to keep cool from the heat.

5 0

3 years ago

PLEASE ANSWER CORRECTLY! MARKING BRANLIEST!

seropon [69]

Answer: oil spill, forest fire

Explanation:

The short term environmental change is a change that occur in an environment which exerts it's effect over the environment and the living beings for a short duration. The damage caused by the short term environmental change can be recovered in a short time. The oil spill and forests fire are the example of short term environmental changes.

The oil spill will reduce the abundance of the aquatic species until the oil is circulated out at the bay of the water body therefore, it is a short term environmental change.

Forests fire will remain ignited for a short period but can be destructive enough to cause damage to flora and fauna species. As time passes the reduced populations of flora and fauna can be recovered again.

5 0

2 years ago

Read 2 more answers

g For the reaction Ag2S(s)⇌2Ag+(aq)+S2−(aq)Ag2S(s)⇌2Ag+(aq)+S2−(aq), Keq=2.4×10−4Keq=2.4×10−4, and the equilibrium concentration

Roman55 [17]

Answer:

0.32 M

Explanation:

Step 1: Write the balanced reaction at equilibrium

Ag₂S(s) ⇌ 2 Ag⁺(aq) + S²⁻(aq)

Step 2: Calculate the concentration of Ag⁺ at equilibrium

We will use the formula for the concentration equilibrium constant (Keq), which is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species.

Keq = [Ag⁺]² × [S²⁻]

[Ag⁺] = √{Keq / [S²⁻]}

[Ag⁺] = √{2.4 × 10⁻⁴ / 0.0023} = 0.32 M

8 0

2 years ago

It has been suggested that the surface melting of ice plays a role in enabling speed skaters to achieve peak performance. Carry

Fittoniya [83]

Answer:

a) Pf = 689.4 bar

b) P = 226.6 bar

c) T = 269.99 K

Explanation:

a)

The molar volume of ice is equal to:

Vi = m/p = (18.02x10^-3 kg H2O/1 mol H2O)*(1 m^3/920 kg) = 1.96x10^-5 m^3 mol^-1

The molar volume of liquid water is equal to:

Vl = (18.02x10^-3 kg/1 mol)*(1 m^3/997 kg) = 1.8x10^-5 m^3 mol^-1

The change in volume is equal to:

ΔVchange = Vi-Vl = 1.96x10^-5 - 1.8x10^-5 = 1.5x10^-6 m^3 mol^-1

using the Clapeyron equation:

Pf = Pi + ((ΔHf*ΔT)/(ΔVf*Ti)) = 1.013x10^5 Pa + ((6010 J mol^-1 * 4.7 K)/(1.5x10^-6 * 273.15 K)) = 6.89x10^7 Pa = 689.4 bar

b)

For the pressure we will use the equation:

P = (m*g)/A, where m is the mass, g is the acceleration of gravity and A is the area. Replacing values:

P = (79 kg * 9.81 m s^-2)/(1.9x10^-4 m * 0.18 m) = 2.26x10^7 Pa = 226.6 bar

c)

From Clapeyron´s expression we need to clear ΔT:

ΔT = ((Pf-Pi)*ΔV*Ti)/ΔHf = ((2.26x10^7 - 6.89x10^7)*1.5x10^-6*273.15)/6010 = -3.16 K

you can evaluate the new melting point of ice:

T = Ti + ΔT = 273.15 K - 3.16 K = 269.99 K

4 0

3 years ago

Complete and balance the chemical equations for the precipitation reactions, if any, between the following pairs of reactants, a

Crank

Explanation:

a. $Pb(NO_3)_2(aq) + Na_2SO_4(aq)$ → ?

$Pb(NO_3)_2(aq) + Na_2SO_4(aq)\rightarrow PbSO_4(s)+2NaNO_3(aq)$

$Pb(NO_3)_2(aq)\rightarrow Pb^{2+}(aq)+2NO_3^{-}(aq)$

$Na_2SO_4(aq)\rightarrow 2Na^++SO_4^{2-}(aq)$

$Pb^{2+}(aq)+2NO_3^{-}(aq)+2Na^++SO_4^{2-}(aq)\rightarrow PbSO_4(s)+2Na^++2NO_3^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)$

b. $NiCl_2(aq) + NH_4NO_3(aq)$ →

$NiCl_2(aq) + NH4NO_3(aq) \rightarrow Ni(NO_3)_2+NH_4Cl(aq)$

No precipitation is occuring.

c. $Fe_Cl2(aq) + Na_2S(aq)$ →

$FeCl_2(aq) + Na_2S(aq)\rightarrow FeS(s)+2NaCl(aq)$

$FeCl_2(aq)\rightarrow Fe^{2+}(aq)+2Cl^{-}(aq)$

$Na_2S(aq)\rightarrow 2Na^++S{2-}(aq)$

$Fe^{2+}(aq)+2Cl^{-}(aq)+2Na^++S^{2-}(aq)\rightarrow FeS(s)+2Na^++2Cl^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Fe^{2+}(aq)+S^{2-}(aq)\rightarrow FeS(s)$

d. $MgSO_4(aq) + BaCl_2(aq)$ →

$MgSO4(aq) + BaCl2(aq)\rightarrow BaSO_4(s)+MgCl_2$

$MgSO_4(aq)\rightarrow Mg^{2+}(aq)+SO_4^{2-}(aq)$

$BaCl_2(aq)\rightarrow Ba^{2+}+2Cl^{-}(aq)$

$Mg^{2+}(aq)+SO_4^{2-}(aq)+Ba^{2+}+2Cl^{-}(aq)(aq)\rightarrow BaSO_4(s)+Mg^{2+}(aq)+2Cl^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

5 0

2 years ago