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3 years ago

All molecules are made up of: A)hydrogen B)minerals C)atoms D)gas

2 answers:

defon3 years ago

6 0

The only answer you can pick from that group is atoms, but it is a shaky answer. It's just that nothing else is appropriate. Hydrogen won't work C-Cl4 is a molecule. No hydrogen.

Minerals. Again CCl4 does not contain any minerals.

Gas: Doesn't work. Many things (perhaps even the majority of compounds) are not gaseous. For example NaCl is table salt. It's a solid

Answer: Choose C. Ions are what make me hesitant.

masya89 [10]3 years ago

4 0

answer c atoms

i think that is correct

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A mixture of copper and tin would be called _________.

Alexus [3.1K]

It would be called an alloy.

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4 years ago

A _________ is an oxidizing agent that, when mixed with an oxidation hair color, supplies the necessary oxygen gas to develop co

Nezavi [6.7K]

Answer:

Developer

Explanation:

A <u><em>developer </em></u>is an oxidizing agent that, when mixed with an oxidation hair color, supplies the necessary oxygen gas to develop color molecules and create a change in hair color.

Mostly hair colors are used to dye the gray hairs. Hair colors are mixed with developer which is act as oxidizing agent. The wisely used developer is hydrogen peroxide. The hair color is made of ammonia and oxidative tints.

These tints consist of small precursors of uncolored dye. These combined with hydrogen peroxide and form large tint molecule. These are then diffuse into cortex of hair and can not get off during shampoo.

These are termed as permanent hair colors.

3 0

4 years ago

5. What is the distance of something from side to side?

Sav [38]

Answer:

Distance of something it can be any type of distance

6 0

3 years ago

A non-mechanical, rigid, and fully insulated mixing tank is used to combine two inflows to produce wet steam. One inflow consist

Bogdan [553]

Explanation:

As it is given that stream A is super heated. From stream tables, we get that specific enthalpy, ( $h_{A}$ ) is 3054.29 kJ/kg.

For stream B, it is saturated water at 25 degree celsius. It's $h_{B}$ is 2546.54 kJ/kg.

Stream C, data will be as follows.

P = 200 kPa, $\chi$ = 0.9

So, $h_{c}$ = $h_{f} + \chi \times h_{fg}$

= 504.47 + 0.9 \times 2201.7

= 2486 kJ/kg

Now, energy balance formula will be as follows.

$m_{A}h_{A} + m_{B}h_{B}$ = $(m_{A} + m_{B})h_{c}$

$3 \times 3054.29 + m_{B} \times 2546.54$ = $(3 + m_{B}) \times 2486$

$m_{B}$ = 28.16 kg/s

Hence, inflow of saturated liquid is 28.16 kg/s. According to steam table, temperature of wet steam is $120.23^{o}C$

Mass flow rate of out flow is 31.16 kg/s.

7 0

4 years ago

Find the enthalpy of neutralization of HCl and NaOH. 87 cm3 of 1.6 mol dm-3 hydrochloric acid was neutralized by 87 cm3 of 1.6 m

Vesna [10]

Answer : The correct option is, (A) -101.37 KJ

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.1392 mole of HCl neutralizes by 0.1392 mole of NaOH

Thus, the number of neutralized moles = 0.1392 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $87ml+87ml=174ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 174ml=174g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 174 g

$T_{final}$ = final temperature of water = 317.4 K

$T_{initial}$ = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

$q=174g\times 4.18J/g^oC\times (317.4-298)K$

$q=14110.008J=14.11KJ$

Thus, the heat released during the neutralization = -14.11 KJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -14.11 KJ

n = number of moles used in neutralization = 0.1392 mole

$\Delta H=\frac{-14.11KJ}{0.1392mole}=-101.37KJ/mole$

Therefore, the enthalpy of neutralization is, -101.37 KJ

3 0

3 years ago