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3 years ago

Give two examples of activities performed by the veterinarian

Chemistry

1 answer:

AURORKA [14]3 years ago

8 0

Answer:

exp:

two ways of activities performed by a veterinarian is:

performing surgery and checkups.

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rite stepwise equations for protonation or deprotonation of this polyprotic species in water. What are the products of the first

adoni [48]

<u>Answer:</u> The products of the given chemical equation are $HCO_3^-\text{ and }OH^-$

<u>Explanation:</u>

Protonation equation is defined as the equation in which protons get added in the substance.

The chemical equation for the protonation of carbonate ion in the presence of water follows:

$CO_3^{2-}+H_2O\rightarrow HCO_3^-+OH^-$

By Stoichiometry of the reaction:

1 mole of carbonate ion reacts with 1 mole of water to produce 1 mole of hydrogen carbonate ion and 1 mole of hydroxide ion

Hence, the products of the given chemical equation are $HCO_3^-\text{ and }OH^-$

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3 years ago

Need help with these two pls!!

denis23 [38]

what is a fire piston?

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3 years ago

Chemical weathering processes do not alter the chemical structure of minerals in rocks. *

Anna007 [38]

I THINK that that is true. It's just weathering the rock down not change composition. Lemme know if you get it right please! (:

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3 years ago

Burka [1]

Answer:

$\large \boxed{\text{E) 721 K; B) 86.7 g}}$

Explanation:

Question 7.

We can use the Combined Gas Laws to solve this question.

a) Data

p₁ = 1.88 atm; p₂ = 2.50 atm

V₁ = 285 mL; V₂ = 435 mL

T₁ = 355 K; T₂ = ?

b) Calculation

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}$

Question 8. I

We can use the Ideal Gas Law to solve this question.

pV = nRT

n = m/M

pV = (m/M)RT = mRT/M

a) Data:

p = 4.58 atm

V = 13.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 385 K

M = 46.01 g/mol

(b) Calculation

$\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}$

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3 years ago

______ is the genetic DNA makeup of the organism. Example: Hh

Hunter-Best [27]

The answer would be A. Genotype

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3 years ago

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