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Ivanshal [37]
3 years ago
13

Convert 300g into kg​

Chemistry
1 answer:
leva [86]3 years ago
6 0

Answer:

0.3kg

Explanation:

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f the Ksp for HgBr2 is 2.8×10−14, and the mercury ion concentration in solution is 0.085 M, what does the bromide concentration
goldfiish [28.3K]

Answer:

0.057 M

Explanation:

Step 1: Given data

Solubility product constant (Ksp) for HgBr₂: 2.8 × 10⁻⁴

Concentration of mercury (II) ion: 0.085 M

Step 2: Write the reaction for the solution of HgBr₂

HgBr₂(s) ⇄ Hg²⁺(aq) + 2 Br⁻

Step 3: Calculate the bromide concentration needed for a precipitate to occur

The Ksp is:

Ksp = 2.8 × 10⁻⁴ = [Hg²⁺] × [Br⁻]²

[Br⁻] = √(2.8 × 10⁻⁴/0.085) = 0.057 M

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3 years ago
Patricia roller skates 0.50 kilometers, straight north to the library. Then she walks 0.25 kilometers to her friend's house, whi
fredd [130]
Number 1 is correct.
6 0
4 years ago
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8 0
3 years ago
A gas made up of atoms escapes through a pinhole 0.392 times as fast as Ne gas. Write the chemical formula of the gas.
Hunter-Best [27]

Answer:  The chemical formula of the gas is Xenon.

Explanation:

From Graham's law of effusion rates, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

\frac{rate_1}{rate_2}=\sqrt{\frac{M_2}{M_1}}

Given: Rate of unknown gas = 0.392\times R_{Ne}  

Putting the values in the formula:

\frac{rate_X}{rate_{Ne}}=\sqrt{\frac{M_{Ne}}{M_X}}

0.392=\sqrt{\frac{20.17}{M_X}}

Squaring both sides:

0.154=\frac{20.17}{M_X}

M_X=131g/mol

As Xenon (Xe) has molar mass of 131g/mol, Thus the chemical formula of the gas is Xenon.

5 0
3 years ago
Given the thermochemical equations below, What is the standard heat of formation of CuO(s)? 2 Cu2O(s) + O2(g) ---> 4 CuO(s) ∆
OlgaM077 [116]

-130KJ is the standard heat of formation of CuO.

Explanation:

The standard heat of formation or enthalpy change can be calculated by using the formula:

standard heat of formation of reaction = standard enthalpy of formation of product - sum of enthalpy of product formation

Data given:

Cu2O(s) ---> CuO(s) + Cu(s) ∆H° = 11.3 kJ

2 Cu2O(s) + O2(g) ---> 4 CuO(s) ∆H° = -287.9 kJ

CuO + Cu ⇒ Cu2O (-11.3 KJ)      ( Formation of Cu2O)

When 1 mole Cu20 undergoes combustion 1/2 moles of oxygen is consumed.

Cu20 + 1/2 02 ⇒ 2CuO (I/2 of 238.7 KJ) or 119.35 KJ

So standard heat of formation of  formation of Cu0 as:

Cu + 1/2 02 ⇒ CuO

putting the values in the equation

ΔHf = ΔH1 + ΔH2     (ΔH1 + ΔH2  enthalapy of reactants)

heat of formation = -11.3 + (-119.35)

                            = - 130.65kJ

-130.65 KJ is the heat of formation of CuO in the given reaction.

7 0
3 years ago
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