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sammy [17]
4 years ago
15

What name should be used for the ionic compound Ga2(SO4)3

Chemistry
1 answer:
Pepsi [2]4 years ago
7 0
The above ionic compound is called gallium sulphate

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What is the change in density if a sample goes from 3.21 g/L to 5.43 g/mL?
Step2247 [10]

Answer:

\Delta \rho =2.22 g/mL

Explanation:

Hello,

In this case, since a change in science is widely known to be considered as a subtraction between the the final and initial values of two measured variables and is represented via Δ, here the final density is 5.43 g/mL and the initial one was 3.21 g/mL, therefore, the change in density is:

\Delta \rho=\rho _f-\rho _i\\ \\\Delta \rho=5.43g/mL-3.21g/mL\\\\\Delta \rho =2.22 g/mL

Best regards.

3 0
3 years ago
The temperature of a sample of a substance changes from 10.°C to 20.°C. How many Kelvin does the temperature change?
Vanyuwa [196]

Answer:

10 kelvin

Explanation:

10 c. =283.15 k.

20 c. = 293.15 k.

293.15 - 283.15 = 10

7 0
3 years ago
An obese man was discovered in his air-conditioned hotel room sitting in a chair in front of the television. The air conditioner
Leno4ka [110]

Answer:

It has been approximately 6 hours after death.

Explanation:

This is because between 2-6 hours after death, the body starts becoming stiff from top to bottom, then spreading to the limbs. Since there is only rigor in his upper body, that would mean that with normal temperature and body conditions, it would be 4 or 5 hours after death. But since he is obese and in cold temperature, there is slower progression of rigor, leading to the maximum time in the first rigor mortis phase, 6 hours.

8 0
3 years ago
½O2(g) + H2(g) ⇌ H2O(g)
ira [324]

Answer:

-241.826 kJ·mol⁻¹;  -146.9 J·K⁻¹mol⁻¹; 664.6 J·K⁻¹mol⁻¹; spontaneous

Explanation:

                        ½O₂(g)   +  H₂(g) ⟶ H₂O(g)

ΔHf°/kJ·mol⁻¹:      0                0        -241.826

S°/J·K⁻¹mol⁻¹:   205.0         130.6       188.7

1. ΔᵣH

ΔᵣH = products -reactants = -241.826 -(0 + 0) = -241.826 kJ·mol⁻¹

2. ΔᵣS

ΔᵣS = products - reactants = 188.7 - (205.0 + 130.6) = 188.7 - 335.6 = -146.9 J·K⁻¹mol⁻¹

3. ΔS(univ)

\begin{array}{rcl}\Delta S_{\text{univ}} &=& \Delta S_{\text{sys}}  +\Delta S_{\text{surr}}\\\\ &=& \Delta S_{\text{sys}}  -\dfrac{\Delta H_{\text{sys}}}{T}\\\\& = & -146.9 - \dfrac{-241826}{298}\\\\& = & -146.9 + 811.5\\& = & \mathbf{664.6 \,\, J\cdot K^{-1}mol^{-1}}\\\end{array}

4. Spontaneity

\begin{array}{rcl}\Delta G &=& \Delta H - T\Delta S\\& = & -241.826 - 298 \times (-0.1469)\\& = & -241.826 + 43.776\\& = &  \textbf{-198.050 kJ}\cdot\textbf{mol}^{\mathbf{-1}}\\\end{array}

ΔG is negative, so the reaction is spontaneous.

4 0
3 years ago
If trees loose there leaves as each seasons progress which biome is that
Kobotan [32]

Answer:

Temperate Deciduous Fores

Explanation:

4 0
3 years ago
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