How do I do this? It's in Chemistry, and it has to do with Single Replacement

What do a mole of magnesium and a mole of iron have in common

lukranit [14]

What they have in common is that they both have the same number of atoms. Hope this helps. :)

7 0

4 years ago

Pls help!!<br>plz answer correctly!<br>will give the brainliest!<br>Urgent!!

blagie [28]

a)

A: Copper

B: CuO

C: $\mathrm{CuSO_4}$

D: $\mathrm{CuCO_3}$

E: $\mathrm{CO_2}$

F: $\mathrm{Cu(NO_3)_2}$

b)

$\mathrm{CuO+ H_2SO_4}\rightarrow \mathrm{CuSO_4 + H_2O}$

c)

$\mathrm{CuCO_3+ 2HNO_3}\rightarrow \mathrm{Cu(NO_3)_2+ CO_2+ H_2O}$

7 0

3 years ago

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Compound X has a molar mass of 416.48 g mol

bazaltina [42]

Answer:

P₂Cl₁₀

Explanation:

From the question given above, the following data were obtained:

Molar mass of compound X = 416.48 g/mol

Percentage of phosphorus (P) = 14.87%

Percentage of Chlorine (Cl) = 85.13%

Molecular formula of X =?

Next, we shall determine the empirical formula of compound X. This can be obtained as follow:

P = 14.87%

Cl = 85.13%

Divide by their molar mass

P = 14.87 / 31 = 0.480

Cl = 85.13 / 35.5 = 2.398

Divide by the smallest

P = 0.480 / 0.480 = 1

Cl = 2.398 / 0.480 = 5

Empirical formula of compound X is PCl₅

Finally, we shall determine the molecular formula of compound X. This can be obtained as follow:

Molar mass of compound X = 416.48 g/mol

Empirical formula = PCl₅

Molecular formula =?

Molecular formula= [Empirical formula]ₙ

[PCl₅]ₙ = 416.48

[31 + (35.5 × 5)]ₙ = 416.48

[31 + 177.5]n = 416.48

208.5n = 416.48

Divide both side by 208.5

n = 416.48 / 208.5

n = 2

Molecular formula = [PCl₅]ₙ

Molecular formula = [PCl₅]₂

Molecular formula = P₂Cl₁₀

Therefore, the molecular formula of compound X is P₂Cl₁₀

4 0

3 years ago

Lithium-6 has a mass of 6.0151 amu and lithium-7 has a mass of 7.0160 amu. The relative abundance of Li-6 is 7.42% and the relat

horsena [70]

Average atomic mass is the weighted average atomic masses with regard to the relative abundance of the isotopes
average atomic mass of Li = relative abundance of Li-6 x mass of Li-6 + relative abundance of Li-7 x mass of Li-7
average atomic mass of Li = (7.42% x 6.0151 a.m.u) + (92.58% x 7.0160 a.m.u)
= 0.446 + 6.495
= 6.941 amu
average atomic mass of Li is 6.941 amu

3 0

3 years ago

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8. The standard enthalpy of formation of RbF(s) is –557.7kJ/mol and the standard enthalpy of formation of RbF(aq, 1 m) is –583.8

garri49 [273]

Explanation:

Given

The enthalpy of formation of RbF (s) is –557.7kJ/mol

The standard enthalpy of formation of RbF (aq, 1 m) is –583.8 kJ/mol

The enthalpy of solution of RbF = Enthalpy of RbF (aq) - Enthalpy of formation of RbF (s)

= -583.8 - (-557.7) kJ/mol

= -26.1 kJ/mol

The enthalpy is negative which means that the temperature will rise when RbF is dissolved.

3 0

3 years ago