Oran fills in the table below to organize information about the gas laws.

Rasek [7]

Its total charge is zero but for the elements:
Sn===> Sn4+ positive
Cl===> Cl- negative

7 0

1 year ago

The ship that will transport the terranauts to the core is built of what material?A. Cobalt B. Diamond C. Kryptonite D. Unobtain

Veseljchak [2.6K]

Answer:

Explanation:

Ketcher 01232019462D 1 1.00000 0.00000 0 5 4 0 0 0 999 V2000 -0.0330 2.2250 0.0000 C 0 0 0 0 0 0 0 0 0 0 0.8330 2.7250 0.0000 C 0 0 0 0 0 0 0 0 0 0 1.6990 2.2250 0.0000 C 0 0 0 0 0 0 0 0 0 0 0.8330 3.7250 0.0000 C 0 0 0 0 0 0 0 0 0 0 1.6990 1.2250 0.0000 C 0 0 0 0 0 0 0 0 0 0 1 2 1 0 0 0 2 3 1 0 0 0 2 4 1 0 0 0 3 5 1 0 0 0 M END

4 0

2 years ago

7.00g of Compound X with molecular formula C5H10 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25

horsena [70]

Answer:

The standard heat of formation of Compound X at 25°C is -3095.75 kJ/mol.

Explanation:

Mass of compound X = 7.00 g

Moles of compound X = $\frac{7.00 g}{70 g/mol}=0.100 mol$

Mass of water in calorimeter ,m= 35.00 kg = 35000 g

Change in temperature of the water in calorimeter = ΔT

ΔT = 2.113°C

Specific heat capacity of water ,c= 4.186 J/g °C

Q = m × c × ΔT

$Q=35000 g\times 4.186 J/g ^oC\times 2.113^oC=309,575.6 J=309.575 kJ$

Heat gained by 35 kg of water is equal to the heat released on burning of 0.100 moles of compound X.

Heat of formation of Compound X at 25°C:

$\frac{-Q}{\text{moles of compound X}}=\frac{-309.575 }{0.100 mol}$

= -3095.75 kJ/mol

6 0

1 year ago

Draw all of the constitutional isomers of the molecule with formula C3H5Br. Ignore geometric and stereoisomers.

Kay [80]

Answer:

-) 3-bromoprop-1-ene

-) 2-bromoprop-1-ene

-) 1-bromoprop-1-ene

-) bromocyclopropane

Explanation:

In this question, we can start with the I.D.H (hydrogen deficiency index):

$I.D.H~=~\frac{(2C)+2+(N)-(H)-(X)}{2}$

In the formula we have 3 carbons, 5 hydrogens, and 1 Br, so:

$I.D.H~=~\frac{(2*3)+2+(0)-(5)-(1)}{2}~=~1$

We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.

We can start with a linear structure with 3 carbon with a double bond in the first carbon and the Br atom also in the first carbon (1-bromoprop-1-ene). In the second structure, we can move the Br atom to the second carbon (2-bromoprop-1-ene), in the third structure we can move the Br to carbon 3 (3-bromoprop-1-ene). Finally, we can have a cyclic structure with a Br atom (bromocyclopropane).

See figure 1

I hope it helps!

7 0

1 year ago

Given that the rate constant is 4.0×10−4 m−1 s−1 at 25.0 ∘c and that the rate constant is 2.6×10−3 m−1 s−1 at 42.4 ∘c, what is t

stepladder [879]

So here's how you find the answer:

Given: (rate constants)

K₁ = 4.0 x 10⁻⁴ M⁻¹s⁻¹.
T₁ = 25.0 C = 293 K.
k₂ = 2.6 x10⁻³ M⁻¹s⁻¹.
T₂ = 42.4 C = 315.4 K.
R = 8,314 J/K·mol.

Use the equation:

ln(k₁/k₂) = Ea/R (1/T₂ - 1/T₁).

Transpose:

Ea = R·T₁·T₂ / (T₁ - T₂) · ln(k₁/k₂)

Substitute within the given transposed equation:

Ea = 8,314 J/K·mol · 293 K · 315.4 K ÷ (293 K - 315.4 K) · ln (4.0 x 10⁻⁴ M⁻¹s⁻¹/ 2.6 x 10⁻³ M⁻¹s⁻¹).

Continuing the solution we get:

Ea = 768315 J·K/mol ÷ (-22,4 K) * (-1.87)

The value of EA is:

Ea = 64140.58 J/mol ÷ 1000 J/kJ = 64.140 kJ/mol.

4 0

1 year ago