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2 years ago

1) Is coffee hot?

Chemistry

2 answers:

Ksivusya [100]2 years ago

4 0

Answer:

1. not testable

2. testable

3. not testable

4. testable

Sav [38]2 years ago

3 0

Answer:

1. Testable

2. Testable

3. I’d say surfing (just my personal opinion)

4. Testable

Explanation:

I guessed

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In the following net ionic equation, identify each reactant as either a Bronsted-Lowry acid or a Bronsted-Lowry base. HCN(aq) H2

vfiekz [6]

Answer:

Explanation:

The definition of acids and bases by Arrhenius Theory was modified and extended by Bronsted-Lowry.

Bronsted-Lowry defined acid as a molecule or ion which donates a proton while a base is a molecule or ions that accepts the proton. This definition can be extended to include acid -base titrations in non-aqueous solutions.

In this theory, the reaction of an acid with a base constitutes a transfer of a proton from the acid to the base.

From the given information:

$\mathsf{HCN _{(aq)} + H_2O_{(l)} \to CN^{-}_{(aq)} + H_3O_{(aq)}}$

From above:

We will see that HCN releases an H⁺ ion, thus it is a Bronsted-Lowry acid

$H_2O$ accepts the H⁺ ion ,thus it is a Bronsted-Lowry base.

The formula of the reactant that acts as a proton donor is <u>HCN</u>

The formula of the reactant that acts as a proton acceptor is <u>H2O</u>

8 0

3 years ago

what temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution (specific heat capacity = 3

ivolga24 [154]

Answer:

1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.

Explanation:

Mass of ethylene glycol = m = 100 g

Specific heat capacity of ethylene glycol = c = 3.5 J/g°C

Change in temperature of ethylene glycol = ΔT

Heat loss by the ethylene glycol = Q = 350 J

$Q=mc\Delta T$

$\Delta T=\frac{Q}{mc}=\frac{350 J}{100 g\times 3.5 J/g^oC}$

ΔT = 1°C

1°C temperature change will be observed if a sample of 100 g of ethylene glycol antifreeze solution.

3 0

3 years ago

Question 231.15 pts A 0.200 M solution of a weak monoprotic acid HA is found to have a pH of 3.00 at room temperature. What is t

MariettaO [177]

Answer:

The correct answer is: Ka= 5.0 x 10⁻⁶

Explanation:

The ionization of a weak monoprotic acid HA is given by the following equilibrium: HA ⇄ H⁺ + A⁻. At the beginning (t= 0) we have 0.200 M of HA. Then, a certain amount (x) is dissociated into H⁺ and A⁻, as is detailed in the following table:

HA ⇄ H⁺ + A⁻

t= 0 0.200 M 0 0

t -x x x

t= eq 0.200M -x x x

At equilibrium, we have the following ionization constant expression (Ka):

Ka= $\frac{ [H^{+}] [A^{-} ]}{ [HA]}$

Ka= $\frac{x x}{0.200 M -x}$

Ka= $\frac{x^{2} }{0.200 M - x}$

From the definition of pH, we know that:

pH= - log [H⁺]

In this case, [H⁺]= x, so:

pH= -log x

3.0= -log x

⇒x = 10⁻³

We introduce the value of x (10⁻³) in the previous expression and then we can calculate the ionization constant Ka as follows:

Ka= $\frac{(10^{-3})^{2} }{0.200 - (10^{-3}) }$ = $\frac{10^{-6} }{0.199}$ = 5.025 x 10⁻⁶= 5.0 x 10⁻⁶

5 0

3 years ago

Read 2 more answers

When .080 moles of propane burn at STP, what volume of carbon dioxide is produced?

ahrayia [7]

Taking into account the reaction stoichiometry and the definition of STP, 5.4 L of carbon dioxide is produced.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

C₃H₈: 1 mole
O₂: 5 moles
CO₂: 3 moles
H₂O: 4 moles

<h3>Moles of CO₂ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 1 mole of C₃H₈ form 3 moles of CO₂, 0.080 moles of C₃H₈ form how many moles of CO₂?

$amount of moles of CO_{2} =\frac{0.080 moles of C_{3} H_{8}x3 moles of CO_{2} }{1 mole of C_{3} H_{8} }$

<u><em>amount of moles of CO₂= 0.24 moles</em></u>

<h3>Definition of STP condition</h3>

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

<h3>Volume of CO₂ produced</h3>

You can apply the following rule of three: If by definition of STP 1 mole of CO₂ occupies 22.4 L, 0.24 moles of CO₂ how much volume does it occupy?

$volume of CO_{2} =\frac{0.24 molesx22.4 L}{1 mole}$