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2 years ago

HELP!! BRAINLIEST 50 POINTS

Chemistry

1 answer:

zhenek [66]2 years ago

6 0

Answer:

The given equation obey the law of conservation of mass.

Explanation:

Chemical equation:

2LiOH + CO₂ → Li₂CO₃ + H₂O

There are equal number of atoms of oxygen, hydrogen and lithium on both side of equation so it obey the law of conservation of mass.

Law of conservation of mass:

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

2LiOH + CO₂ → Li₂CO₃ + H₂O

2(6.941 + 16 + 1) + 12+32 6.941×2 + 12 + 3×16 + 18

47.882 + 44 13.882 +12+48 + 18

91.882 g 91.882 g

The mass of reactants and product are equal.

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List of 10 strong bases and acids?

tatyana61 [14]

KaAcid1.0 * 109Hydrobromic acidHBr1.3 * 106Hydrochloric acidHCl1.0 * 103Sulfuric acidH2SO42.4 * 101Nitric acidHNO3

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3 years ago

In an accident, a solution containing 2.5 kg of nitric acid was spilled. Two kilograms of Na2CO3 was quickly spread on the area

storchak [24]

First we must write a balanced chemical equation for this reaction

$Na_2CO_3 _(_a_q_)+ 2HNO_3_(_a_q_) \implies 2NaNO_3_(_s_) + CO_2_(_g_) + H_2O_(_l_)$

The mole ratio for the reaction between $HNO_3$ and $Na_2CO_3$ is 1:2. This means 1 moles of $Na_2CO_3$ will neutralize 2 moles $HNO_3$ . Now we find the moles of each reactant based on the mass and molar mass.

$2500g HNO_3 \times \frac{mol}{63.01g\ HNO_3} = 39.67 mol\ HNO_3$

$2000g\ Na_2CO_3 \times \frac{mol}{105.99g \ Na_2CO_3} = 18.87 mol\ Na_2CO_3$

$\frac{18.87 mol Na_2CO_3}{39.67\ HNO_3} = \frac{1 molNa_2CO_3}{2 mol HNO_3}$

The $Na_2CO_3$ was enough to neutralize the acid because 18.87:39.67 is the same as 1:2 mol ratio.

4 0

3 years ago

Equal volumes of SO2(g) and O2(g) at STP contain the same number of

LuckyWell [14K]

Answer:

Equal volumes of SO2(g) and O2(g) at STP contain the same number of molecules

Explanation:

According to Avogadro Law,

Equal volume of all the gases at same temperature and pressure have equal number of molecules.

This law state that volume and number of moles of gas have direct relation.

When the amount of gas increases its volume will increase and when the amount of gas decreases its volume will decrease.

Mathematical relation:

V ∝ n

V/n = K

K is proportionality constant.

When number of moles change from n₁ to n₂ and volume from V₁ to V₂

expression will be,

V₁/n₁ = K , V₂/n₂ = K

V₁/n₁ = V₂/n₂

8 0

2 years ago

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What is the missing number in the sequence 3,5,9,_,15,23,33

balu736 [363]

Answer:

In every sequence even numbers are added in order.

3, 5, 9, 15, 23, 33

6 0

2 years ago

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Write the oxidation and reduction half reactions;

luda_lava [24]

Answer:

$Fe^{2+}$ ⇒ $Fe^{3+}+e^-$

$Br_2+2e^-$ ⇒ $2Br^-$

$Mg$ ⇒ $Mg^{2+}+2e^-$

$Cr^{3+}+e^-$ ⇒ $Cr^{3+}$

Explanation:

Remember that positive number superscripts mean electrons lack and negative numbers mean electrons 'excess' (if we compare it with the neutral element). So, for the case of Fe2+ which is converted to Fe3+, we know that in Fe2+ there is a two electrons lack, while in Fe3+ there is a 3 electrons lack; it means that Fe2+ was converted to Fe3+ but releasing one electron:

$Fe^{2+}$ ⇒ $Fe^{3+}+e^-$

The same analysis is applied to Br2; Br2 is a molecule which is said to have a zero superscript because it is an apolar covalent bond; and it is converted to Br-, which, according to what I wrote above, means that there is a one electron excess. So, Br2 must have received an electron in order to change to Br-; but Br2 can't change to Br- as simple as that because Br2 is a molecule, not an atom; it is a molecule that has two Br atoms, so, Br2 must give two Br- ions as products, but receiving one electron for each one:

$Br_2+2e^-$ ⇒ $2Br^-$

Applying the same, in Mg2+ there is a 2 electrons lack, and in Mg is not electron lack (its superscript is zero), so Mg must have released two electrons in order to change to Mg2+:

$Mg$ ⇒ $Mg^{2+}+2e^-$

Cr3+ has a 3 electrons lack, and Cr2+ a two electrons one, so, Cr3+ must receive an electron to convert to Cr2+:

$Cr^{3+}+e^-$ ⇒ $Cr^{3+}$

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3 years ago

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