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3 years ago

Define monocarboxlic acid

Chemistry

1 answer:

Ann [662]3 years ago

4 0

Answer:

When a molecule has just has one carboxylic acid group, it is defined as a monocarboxylic acid.

Explanation:

hope it will help ^_^

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Nora's family went camping. Which way did her family MOST LIKELY produce heat energy while camping? A. By burning B. By friction

strojnjashka [21]

Answer:

B friction is where you rub your something togother and it heats it up so B

Explanation:

6 0

3 years ago

What is the change in temperature in a 128g sample of titanium if it absorbs 2808J of heat energy at a temperature of 2°C? The s

Katena32 [7]

Given:

128g sample of titanium

2808J of heat energy

specific heat of titanium is 0.523 J/ g °C.

Required:

Change in temperature

Solution:

This can be solved through the equation H = mCpT where H is the heat, m is the mass, Cp is the specific heat and T is the change in temperature.

Plugging in the values into the equation

H = mCpT

2808J = (128g) (0.523 J /g °C) T

T = 41.9 °C

3 0

3 years ago

Using the periodic table entry below, match the phrases with their corresponding values.

zhannawk [14.2K]

Atomic mass = 39

number of neutrons = 20

Number of protons= 19

Electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

The number of electron in first shell = 2

number of electrons in n=3 shell = 8

5 0

3 years ago

Read 2 more answers

Whats the correct answer ???

pogonyaev

Which question did you want me to answer

4 0

3 years ago

The common titanium alloy known as T-64 has a composition of 90 weight% titanium 6 wt% aluminum and 4 wt% vanadium. Calculate th

Anna007 [38]

Explanation:

Suppose in 100 g of alloy contains 90% titanium 6% aluminum and 4% vanadium.

Mass of titanium = 90 g

Moles of titanium = $\frac{90 g}{47.87 g/mol}=1.8800 mol$

Total number of atoms of titanium , $a_t=1.8800 mol\times N_A$

Mass of aluminum = 6 g

Moles of aluminium = $\frac{6 g}{26.98 g/mol}=0.2223 mol$

Total number of atoms of aluminium, $a_a=0.2223 mol\times N_A$

Mass of vanadium = 4 g

Moles of vanadium= $\frac{4 g}{50.94 g/mol}=0.0785 mol$

Total number of atoms of vanadium $a_v=0.0785 mol\times N_A$

Total number of atoms in an alloy = $a_t+a_a+a_v$

Atomic percentage:

$Atomic\%=\frac{\text{total atoms of element}}{\text{Total atoms in alloy}}\times 100$

Atomic percentage of titanium:

: $\frac{a_t}{a_t+a_a+a_v}\times 100=\frac{1.8800 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=86.20\%$

Atomic percentage of Aluminium:

: $\frac{a_a}{a_t+a_a+a_v}\times 100=\frac{0.2223 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=10.19\%$

Atomic percentage of vanadium

: $\frac{a_v}{a_t+a_a+a_v}\times 100=\frac{0.0785 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=3.59\%$

6 0

4 years ago