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3 years ago

What type of atoms typically form covalent bonds

Chemistry

1 answer:

d1i1m1o1n [39]3 years ago

3 0

Answer:

Covalent bonds usually occur between nonmetals.

Explanation:

For example, in water (H2O) each hydrogen (H) and oxygen (O) share a pair of electrons to make a molecule of two hydrogen atoms single bonded to a single oxygen atom.

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Which of these statements is not true about chemical reaction rates? a Temperature can change a reaction rate. b The amount of r

Umnica [9.8K]

Answer:

The statement that is not true is: 'Temperature does not affect the reaction rate'.

Explanation:

a) Temperature can change a reaction rate. This is true

Increasing the temperature increases the reaction rates because of the disproportionately large increase in the number of high energy collisions. It is only these collisions (possessing at least the activation energy for the reaction) which result in a reaction.

For example, the time taken to melt a metal will be much higher at a lower temperature but it will decrease as soon as we increase the temperature

b) The amount of reactants can increase the reaction rate. This is true

A higher concentration of reactants leads to more effective collisions per unit time, which leads to an increased reaction rate.

c) Temperature can decrease the reaction rate. This is true

Decreasing the temperature decreases the reaction rates because of the decrease in the number of high energy collisions. It will result in a slower reaction.

d) Temperature does not affect the reaction rate. This is not true.

The reaction rate is temperature dependent. The reaction rate increases with higher temperature and decreases with lower temperature.

4 0

3 years ago

Name the following alkane:

vladimir1956 [14]

The names of the alkanes are 2,2- dimethylbutane, 2,4- dimethylhexane, 2,2,3,3- tetramethylbutane and 4- ethyl, 3,6- dimethyl heptane.

Alkanes are saturated hydrocarbon in the organic chemistry. These are organic compounds that consists of single bonded carbon and hydrogen atoms. The common formula for writing an alkane is given by CₙH₂ₙ₊₂.

Alkanes are further divided into three more types which are:

1. Chain alkanes

2. Cycloalkanes

3. Branched alkanes

Hydrogenation method is used for preparation of alkanes from alkene and alkyne.

The names of the given compounds are:

Part A:

2,2- dimethyl butane

Part B:

2,4- dimethyl hexane

Part C:

2,2,3,3- tetramethyl butane

Part D:

4- ethyl, 3,6- dimethyl heptane

Learn more about alkanes from the link given below.

brainly.com/question/16837399

#SPJ1

4 0

9 months ago

(URGENT FOR BRAINLIEST!!)

aniked [119]

Answer:

mountain

Explanation:

when plates move towards each other they create mountains

8 0

2 years ago

What is the ph of a buffer prepared by adding 0.809 mol of the weak acid ha to 0.608 mol of naa in 2.00 l of solution? The disso

Anarel [89]

The pH of the buffer is 6.1236.

Explanation:

The strength of any acid solution can be obtained by determining their pH. Even the buffer solution strength of the weak acid can be determined using pH. As the dissociation constant is given, we can determine the pKa value as the negative log of dissociation constant value.

$pKa=-log[H] = - log [ 5.66 * 10^{-7}]\\ \\pka = 7 - log (5.66)=7-0.753=6.247\\\\pka = 6.247$

The pH of the buffer can be known as

$pH = pK_{a} + log[\frac{[A-]}{[HA]}}]$

The concentration of $[A^{-}] = Moles of [A]/Total volume = 0.608/2 = 0.304 M\\$

Similarly, the concentration of [HA] = $\frac{Moles of HA}{Total volume} = \frac{0.809}{2} = 0.404$

Then the pH of the buffer will be

pH = 6.247 + log [ 0.304/0.404]

$pH = 6.247 + log 0.304 - log 0.404=6.247-0.517+0.3936=6.1236$

So, the pH of the buffer is 6.1236.

5 0

3 years ago

at a certain pressure the density of superficial carbon dioxide is 0.469. what is the mass of a 25.0 mL sample of supercritical

Murrr4er [49]

Answer:

$\large \boxed{\text{11.7 g}}$

Explanation:

$\text{Mass} = \text{25.0 mL} \times \dfrac{\text{0.469 g}}{\text{1 mL}} = \textbf{11.7 g}\\\\\text{The mass of the carbon dioxide is $\large \boxed{\textbf{11.7 g}}$}$

7 0

2 years ago