All categories

3 years ago

Which of the following bases can remove a proton from formic acid in a reaction that favors products?

Chemistry

1 answer:

Debora [2.8K]3 years ago

7 0

Hydroxide ion is a strong and would react with H+ to form water

OH-+H+---->H2O

You might be interested in

Which property is NOT used to separate a mixture?

NeTakaya

Answer: ability to undergo a chemical reaction

Explanation:

6 0

3 years ago

Which of the following is a property of carbon atoms?

marusya05 [52]

Answer:

B. They can for covalent bonds with other atoms.

Explanation:

Carbon = King of the elements on the periodic table

For its proclivity to form stable covalent bonds with multivalent atoms.

5 0

3 years ago

Read 2 more answers

Does temperature change during a chemical reaction?

Anarel [89]

Answer: yes

Explanation:

Not all the time, but temperature change can happen during a chemical reaction.

4 0

3 years ago

Read 2 more answers

slader An experiment is carried out where 13.9 g of solid NaOH is dissolved in 250.0 g of water in a coffee-cup calorimeter. Dis

kvv77 [185]

Answer:

The mass of the surrounding is $M_t = 263.9 \ g$

Explanation:

From the question we are told that

The mass of $NaOH$ is $m = 13.9 \ g$

The mass of water is $m_w = 250.0g$

The chemical equation for the dissociation process is

$NaOH _{(s) } ---> Na^{+}_{(aq)} + OH^{-} _{(aq)}$

The specific heat capacity of the mixture is $c_p = 4.18 J g^{-1} C^{-1}$

The combined mass of the solution is

$M_t = 13 + 250$

$M_t = 263.9 \ g$

The mass of the surround here is the mass of the coffee-cup calorimeter and this contain the mixture ( water and the NaOH ) so the mass of the surrounding is

$M_t = 263.9 \ g$

3 0

3 years ago

A sample of nitrogen is initially at a pressure of 1.7 kPa, a temperature of -10 C and a volume of 7.5 m3. Then the volume is de

zhannawk [14.2K]

Answer:

$\boxed{\text{2.6 kPa}}$

Explanation:

To solve this problem, we can use the Combined Gas Laws:

$\dfrac{p_{1}V_{1} }{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}$

Data:

p₁ = 1.7 kPa; V₁ = 7.5 m³; T₁ = -10 °C

p₂ = ?; V₂ = 3.8 m³; T₂ = 200 K

Calculations:

(a) Convert temperature to kelvins

T₁ = (-10 + 273.15) K = 263.15 K

(b) Calculate the pressure

$\begin{array}{rcl}\dfrac{1.7 \times 7.5 }{263.15} & = & \dfrac{p_{2} \times 3.8}{200}\\\\0.0485 & = & 0.0190p_{2}\\p_{2} & = & \textbf{2.6 kPa}\\\end{array}\\\text{The new pressure of the gas is \boxed{\textbf{2.6 kPa}}}$

7 0

3 years ago