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3 years ago

365g MnSiO3 = ___ mol

1 answer:

7 0

2.79 mol of MnSiO3

This will be very close but may not be exactly accurate, depending on what your periodic table says.

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Anthony's homework assignment is to demonstrate that an orange has already undergone a chemical change. Which of the following

wolverine [178]

B. Rotten orange is the correct answer. Hope this helps!

8 0

3 years ago

Read 2 more answers

A solution of acetic acid and water contains 205.0 g L-1 of acetic acid and 820.0 g L-1 of water. Compute the density of the sol

LekaFEV [45]

Answer:

$\rho_t=1025000 gmL^{-1}$

Explanation:

From the question we are told that:

Density of acetic acid $\rho_a=205 gL^{-1}$

Density of Water $\rho_w=820 gL^{-1}$

Generally the equation for Solution Density is mathematically given by

$\rho_t= \rho_w+\rho_a$

$\rho_t=205+820$

$\rho_t=1025 gL^{-1}$

$\rho_t=1025000 gmL^{-1}$

6 0

3 years ago

The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g

ra1l [238]

Answer:

1) When 6.97 grams of sodium(s) react with excess water(l), 56.0 kJ of energy are evolved.

2) When 10.4 grams of carbon monoxide(g) react with excess water(l), 1.04 kJ of energy are absorbed.

Explanation:

1) The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g).

2 Na(s) + 2H₂O(l) ⇒ 2NaOH(aq) + H₂(g) ΔH = -369 kJ

The enthalpy of the reaction is negative, which means that 369 kJ of energy are evolved per 2 moles of sodium. The energy evolved for 6.97 g of Na (molar mass 22.98 g/mol) is:

$6.97g.\frac{1mol}{22.98g} .\frac{-369kJ}{2mol} =-56.0kJ$

2) The following thermochemical equation is for the reaction of carbon monoxide(g) with water(l) to form carbon dioxide(g) and hydrogen(g).

CO(g) + H₂O(l) ⇒ CO₂(g) + H₂(g) ΔH = 2.80 kJ

The enthalpy of the reaction is positive, which means that 2.80 kJ of energy are absorbed per mole of carbon monoxide. The energy evolved for 10.4 g of CO (molar mass 28.01 g/mol) is:

$10.4g.\frac{1mol}{28.01g} .\frac{2.80kJ}{mol} =1.04kJ$

3 0

3 years ago

A 25.0 mL aliquot of 0.0680 M EDTA was added to a 59.0 mL solution containing an unknown concentration of V3 . All of the V3 pre

givi [52]

Answer:

$\mathbf{0.02 M}$

Explanation:

$\text{So, from the given question:}$

$\text{EDTA will make complex with}$ $V^{+3}$ $\text{and the remaining EDTA will react with }$ $Ga^{+3}$

$\text{Hence, the total concentration of}$ $V^{+3}$ & $Ga^{+3}$ $\text{will be equivalent to EDTA concentration.}$

$V_{EDTA} = 25 \ mL$

$V_{V^{+3}} = 59.0 \ mL$

$V_{Ga^{+3}} = 13.0 \ mL$

$M_{EDTA} = 0.0680 \ M$

$M_{V^{+3}} = ???(unknown)$

$M_{Ga^{+3}} = 0.0400 \ M$

$V^{+3} + EDTA \to V[EDTA] + EDTA(Excess) \to^{CoA} \ Ga[EDTA] _{complex}$

$M_{EDTA} \times V_{EDTA} = ( V_{V^+3}\times M_{V^{+3}}+ V_{Ga^{+3} }\times M_{Ga^{+3}}})$

$0.0680 \times 25 = (59\times x + 13 \times 0.040) \\ \\ 1.7 = 59x + 0.52\\ \\ 1.7 - 0.52 = 59x \\ \\ 59x = 1.18$

$x = \dfrac{1.18}{59}$

$\mathbf{x =0.02 \ M }$

5 0

3 years ago

The substance water always has a mass ratio of 11% H to 89% O. If 5.00g of a substance containing H and O was decomposed into .2

Gre4nikov [31]

Answer:

No the substance is not water.

Explanation:

The balance chemical equation for the decomposition of water is as follow;

2 H₂O = 2 H₂ + O₂

Step 1: Calculate moles of H₂O;

Moles = Mass / M.Mass

Moles = 5.0 g / 18.01 g/mol

Moles = 0.277 moles of H₂O

Step 2: Calculate Moles of O₂ and H₂ produced by 0.277 moles of H₂O:

According to equation,

2 moles of H₂O produced = 1 mole of O₂

So,

0.277 moles of H₂O will produce = X moles of O₂

Solving for X,

X = 0.277 mol × 1 mol / 2 mol

X = 0.138 moles of O₂

Also,

According to equation,

2 moles of H₂O produced = 2 mole of H₂

So,

0.277 moles of H₂O will produce = X moles of H₂

Solving for X,

X = 0.277 mol × 2 mol / 2 mol

X = 0.227 moles of H₂

Step 3: Calculate Mass of O₂ and H₂ as;

For O₂:

Mass = Moles × M.Mass

Mass = 0.138 mol × 31.99 g/mol

Mass = 4.44 g of O₂

For H₂:

Mass = Moles × M.Mass

Mass = 0.227 mol × 2.01 g/mol

Mass = 0.559 g of H₂

Conclusion:

From conclusion it is proved that the amount of H₂ produced by decomposition of 5 g of water should be 0.559 g while in statement it is less i.e. 0.290 g.

6 0

3 years ago