What percentage of known species are invertebrates?

In the following equation, which compounds are the reactants?NaCl + AgNO → NaNO + AgCl

tangare [24]

NaCl + AgNO are the reactants since the reactants are always on the left side of the yield.

7 0

3 years ago

A weather balloon was initially at a pressure of 0.900 atm, and its volume was 35.0 L. The pressure decreased to 0.750 atm, with

marusya05 [52]

Boyle's Law

P₁V₁=P₂V₂

0.9 x 35 = 0.75 x V₂

V₂ = 42 L

3 0

2 years ago

Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +

qwelly [4]

Answer: The $\Delta H^o_{rxn}$ for the reaction is -1052.8 kJ.

Explanation:

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

$X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)$ $\Delta H^o_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)$ $\Delta H_1=-668.5kJ$

(2) $XCO_3(s)\rightarrow XO(s)+CO_2$ $\Delta H_2=+384.3kJ$

The expression for enthalpy of the reaction follows:

$\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ$

Hence, the $\Delta H^o_{rxn}$ for the reaction is -1052.8 kJ.

7 0

3 years ago

Be sure to answer all parts. identify and label the species in each reaction. (a) nh4+(aq) + h2o(l) ⇌ nh3(aq) + h3o+(aq) acid ba

quester [9]

A)

NH⁴⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃0⁺(aq)

- acid a species that able to donate (H+): NH⁴⁺
- base a species that is able to accept a proton (H+): H₂O
- conjugate base a species formed when acid donates a proton (H+): NH₃
- conjugate acid a species formed by a base accepts a proton (H+): H₃0⁺

b)

CN⁻(aq) + H₂O(l) ⇌ HCN(aq) + OH⁻(aq)

- base a species that is able to accept a proton (H+): CN⁻
- acid a species that able to donate (H+): H₂O
- conjugate acid a species formed by a base accepts a proton (H+): HCN
- conjugate base a species formed when acid donates a proton (H+): OH⁻

5 0

4 years ago

Consider a sample of 3.5 mol of N2(g) at T1 = 350 K, that undergoes a reversible and adiabatic change in pressure from p1 = 1.50

devlian [24]

Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C $_{V}$ You can work out C $_{V}$

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC $_{V}$ ∆T

Polyatomic gas: C $_{V}$ = 3R

∆U= nC $_{V}$ ∆T

∆U= 28g x C $_{V}$ x (350K - 58.3K)

∆U = 28C $_{V}$ x 291.7

∆U = 10967.6 x C $_{V}$

5 0

3 years ago