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2 years ago

Number 1 please fast answer

Chemistry

2 answers:

dimulka [17.4K]2 years ago

7 0

Answer: Color change, Production of odor, and

Temputure change, formation of bubbles, and form of solids

Explanation:

It basic chemistry , as in if the solution has changed it would express itself in these ways otherwise it would not make the list.

Natasha2012 [34]2 years ago

3 0

Answer:

I will give you 5

Explanation:

color chage, formation of a precipitate, formation of a gas, odor change, temperature change. are all signs of a chemical reaction

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100 level student of the adamawa state university mubi was ask to synthesize ammonia (haber process) in the ratio 3:1 of hydroge

lapo4ka [179]

The equilibrium constant, Kc, can be determined knowing that the balanced chemical reaction is N2 + 3H2 = 2 NH3. At equilibrium, Kc is equal to (concentration of NH3)^2 / [(concentration of N2) x (concentration of H2)^3]. Using the given values, Kc is computed to be equal to 1.9977 x 10^-4.

4 0

2 years ago

uppose you are titrating an acid of unknown concentration with a standardized base. At the beginning of the titration, you read

OverLord2011 [107]

Answer:

18.91 ml

Explanation:

Initial volume of base=2.04 ml

Final volume of base = 20.95 ml

Volume of base used= 20.95 - 2.04 = 18.91 ml

Note that the volume of base used is obtained as the difference between the final and initial volume of base, hence the answer given above.

5 0

2 years ago

1000 cm3 of hydrogen gas (hydrogen molecules, H2) contains x number of molecules at room temperature and pressure. Determine the

Gekata [30.6K]

Answer: Number of atoms in $500cm^3$ of radon gas (radon atoms) at the same temperature and pressure is $\frac{x}{2}$

Explanation:

According to avogadro's law, volume of a gas is directly proportional to the number of moles present when temperature and pressure is constant.:

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

$V_1 = Volume of the hydrogen gas = 1000cm^3$

$n_1$ = moles of hydrogen gas = $\frac{xmolecules}{6.023\times 10^{23}molecules}$

$V_2= Volume of the radon gas =500cm^3$

$n_1$ = moles of radon gas = ?

$\frac{1000}{\frac{x}{6.023\times 10^{23}}}=\frac{500}{\frac{y}{6.023\times 10^{23}}}$

$y=\frac{x}{2}$

Thus the number of atoms in $500cm^3$ of radon gas (radon atoms) at the same temperature and pressure is $\frac{x}{2}$

7 0

2 years ago

What is the net ionic equation for the reaction between aqueous solutions of na3po4 and cuso4?

agasfer [191]

Answer:

3Cu²⁺ + 2PO₄³⁻ → Cu₃(PO₄)₂ (s)

Explanation:

Double displacement reaction: A chemical reaction in which aqueous solution of two salts react to give two products by exchanging their counterions.

The chemical equation for the chemical reaction between Na₃PO₄ (aq) and CuSO₄ (aq) is given as:

2 Na₃PO₄ (aq) + 3 CuSO₄ (aq) → Cu₃(PO₄)₂ (s) + 3 Na₂SO₄ (aq)

The full ionic equation for this chemical reaction is:

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 3SO₄²⁻(aq) → Cu₃(PO₄)₂ (s) + 6Na⁺(aq) + 3SO₄²⁻(aq)

Therefore, the net ionic equation for this chemical reaction is:

3Cu²⁺ + 2PO₄³⁻ → Cu₃(PO₄)₂ (s)

4 0

2 years ago

A gas is contained in a thick-walled balloon. When the pressure changes from 2.95 atm to atm, the volume changes from 7.456 L to

zimovet [89]

Let's assume that the gas in the balloon is an ideal gas.

We can use combined gas law,

PV/T = k (constant)

Where, P is the pressure of the gas, V is volume of the gas and T is the temperature of the gas in Kelvin.

For two situations, we can use that as,

P₁V₁/T₁= P₂V₂/T₂

P₁ = 2.95 atm

V₁ = 7.456 L

T₁ = 379 K

P₂ = ?

V₂ = 4.782 L

T₂ = 212 K

By applying the formula,

2.95 atm x 7.456 L / 379 K = P₂ x 4.782 L / 212 K

P₂ = (2.95 atm x 7.456 L x 212 K) / (379 K x 4.782 L)

P₂ = 2.57 atm

Hence, the answer is 2.57 atm.

3 0

3 years ago