What is this equation balanced?<br> ____H2O + ____SO3 → ___

Which of the following is an example of pseudoscience?

UkoKoshka [18]

Answer:

A) Astronomy

Explanation:

8 0

4 years ago

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If 146.5 g of chromium(lll) chlorate decomposes, what mass of oxygen can be produced?

Vinil7 [7]

Answer:

The Equation for this reaction is as follow,

2 Cr(ClO₃)₃ → 2 CrCl₃ + 9 O₂

According to eq,

604.68 g (2mole) of Cr(ClO₃)₃ produces = 288 g (9 moles) of O₂

So,

146.5 g of Cr(ClO₃)₃ will produce = X g of O₂

Solving for X,

X = (146.5 g × 288 g) ÷ 604.68 g

X = 69.77 g of O₂

8 0

3 years ago

Copper is formed when aluminum reacts with copper (II) sulfate in a single-replacement reaction. How many moles of copper can be

serg [7]

Answer:

The answer to your question is the limiting reactant is CuSO₄ and 0.975 moles of Cu were obtained

Explanation:

moles of Copper = ?

mass of Aluminum = 29 g

mass of CuSO₄ = 156 g

Limiting reactant = ?

Balanced Chemical reaction

3 CuSO₄ + 2 Al ⇒ 3 Cu + Al₂(SO₄)₃

Calculate the moles of reactants

CuSO₄ = 64 + 32 + (16 x 4) = 160g

Al = 27 g

160 g of CuSO₄ ----------------- 1 mol

156 g ----------------- x

x = (156 x 1) / 160

x = 0.975 moles

27 g of Al -------------------------- 1 mol

29 g of Al -------------------------- x

x = (29 x 1)/27

x = 1.07 moles

Calculate proportions to find the limiting reactant

Theoretical 3 moles CuSO₄/2 moles Al = 1.5 moles

Experimental 0.975 moles CuSO₄/1.07 moles = 0.91

The experimental proportion was lower than the theoretical proportion that means that the limiting reactant is CuSO₄.

3 moles of CuSO₄ ------------------ 3 moles of Cu

0.975 moles of CuSO₄ --------------- x

x = (0.975 x 3)/3

x = 0.975 moles of Cu were obtained.

3 0

3 years ago

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How do you know if a equation is balanced or unbalanced

denpristay [2]

A chemical equation is only balanced once there are equal numbers of all types of element on both sides of the equation

7 0

3 years ago

What is the electric force on a proton 2.5 fmfm from the surface of the nucleus? Hint: Treat the spherical nucleus as a point ch

sammy [17]

Explanation:

It is known that charge on xenon nucleus is $q_{1}$ equal to +54e. And, charge on the proton is $q_{2}$ equal to +e. So, radius of the nucleus is as follows.

r = $\frac{6.0}{2}$

= 3.0 fm

Let us assume that nucleus is a point charge. Hence, the distance between proton and nucleus will be as follows.

d = r + 2.5

= (3.0 + 2.5) fm

= 5.5 fm

= $5.5 \times 10^{-15} m$ (as 1 fm = $10^{-15}$ )

Therefore, electrostatic repulsive force on proton is calculated as follows.

F = $\frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}$

Putting the given values into the above formula as follows.

F = $\frac{1}{4 \pi \epsilon_{o}} \frac{q_{1}q_{2}}{d^{2}}$

= $(9 \times 10^{9}) \frac{54e \times e}{(5.5 \times 10^{-15})^{2}}$

= $(9 \times 10^{9}) \frac{54 \times (1.6 \times 10^{-19})^{2}}{(5.5 \times 10^{-15})^{2}}$

= 411.2 N

or, = $4.1 \times 10^{2}$ N

Thus, we ca conclude that $4.1 \times 10^{2}$ N is the electric force on a proton 2.5 fm from the surface of the nucleus.

8 0

4 years ago

What is this equation balanced? ____H2O + ____SO3 → ____H2SO4

What is this equation balanced? H2O + SO3 → H2SO4