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2 years ago

Changing a solid to a liquid to a gas is changing the matter. * Your answer

Chemistry

2 answers:

sukhopar [10]2 years ago

6 0

Answer:

i could help you can you explain more plzzzzzz i really need points

Explanation:

Pachacha [2.7K]2 years ago

6 0

Answer:

It is changing the state of matter.

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Describe the chemical reaction based on the chemical equation below also whether the equation is balanced

Crank

Answer:

Its not balanced

4NH₃ + 7O₂ ⇒ 4NO + 6H₂O

3 0

2 years ago

What is the total number of electrons in a<br> Mg2+ ion?(1) 10 (3) 14<br> (2) 12 (4) 24

Citrus2011 [14]

The answer is (1) 10. The protons of Mg atom is 12. So the Mg atom has 12 electrons. The Mg2+ ion has lost two electrons so it has two positive charge. Then the answer is 10 electrons.

5 0

2 years ago

10 m3 of carbon dioxide is originally at a temperature of 50 °C and pressure of 10 kPa. Determine the new density and volume of

Dimas [21]

Answer : The new density and new volume of carbon dioxide gas is 0.2281 g/L and $7.2m^3$ respectively.

Explanation :

First we have to calculate the new or final volume of carbon dioxide gas.

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 10 kPa

$P_2$ = final pressure of gas = 15 kPa

$V_1$ = initial volume of gas = $10m^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $50^oC=273+50=323K$

$T_2$ = final temperature of gas = $75^oC=273+75=348K$

Now put all the given values in the above equation, we get:

$\frac{10kPa\times 10m^3}{323K}=\frac{15kPa\times V_2}{348K}$

$V_2=7.2m^3$

The new volume of carbon dioxide gas is $7.2m^3$

Now we have to calculate the new density of carbon dioxide gas.

$PV=nRT\\\\PV=\frac{m}{M}RT\\\\P=\frac{m}{V}\frac{RT}{M}\\\\P=\rho \frac{RT}{M}\\\\\rho=\frac{PM}{RT}$

Formula for new density will be:

$\rho_2=\frac{P_2M}{RT_2}$

where,

$P_2$ = new pressure of gas = 15 kPa

$T_2$ = new temperature of gas = $75^oC=273+75=348K$

M = molar mass of carbon dioxide gas = 44 g/mole

R = gas constant = 8.314 L.kPa/mol.K

$\rho$ = new density

Now put all the given values in the above equation, we get:

$\rho_2=\frac{(15kPa)\times (44g/mole)}{(8.314L.kPa/mol.K)\times (348K)}$

$\rho_2=0.2281g/L$

The new density of carbon dioxide gas is 0.2281 g/L

5 0

2 years ago

Calculate the pH at the equivalence point for the titration of 0.110 M methylamine (CH3NH2) with 0.110 M HCl. The Kb of methylam

nikklg [1K]

The reaction between the reactants would be:

CH₃NH₂ + HCl ↔ CH₃NH₃⁺ + Cl⁻

Let the conjugate acid undergo hydrolysis. Then, apply the ICE approach.

CH₃NH₃⁺ + H₂O → H₃O⁺ + CH₃NH₂
I 0.11 0 0
C -x +x +x
E 0.11 - x x x

Ka = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]

Since the given information is Kb, let's find Ka in terms of Kb.

Ka = Kw/Kb, where Kw = 10⁻¹⁴

So,
Ka = 10⁻¹⁴/5×10⁻⁴ = 2×10⁻¹¹ = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
2×10⁻¹¹ = [x][x]/[0.11-x]
Solving for x,
x = 1.483×10⁻⁶ = [H₃O⁺]

Since pH = -log[H₃O⁺],
pH = -log(1.483×10⁻⁶)
<em>pH = 5.83</em>

4 0

3 years ago

What would be the mass of a 33.5dm3 sample of O2 at STP?

liraira [26]

Answer:

• One mole of oxygen is equivalent to 16 grams.

→ But at STP, 22.4 dm³ are occupied by 1 mole.

${ \tt{22.4 \: {dm}^{3} \: \dashrightarrow \: 16 \: grams}} \\ { \tt{33.5 \: {dm}^{3} \: \dashrightarrow \: ( \frac{33.5 \times 16}{22.4} ) \: grams}} \\ \\ \dashrightarrow \: { \boxed{ \tt{23.94 \: g \approx \: 24 \: grams}}}$

7 0

2 years ago