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3 years ago

What is the correct name for N4Cl2

Chemistry

1 answer:

Dahasolnce [82]3 years ago

4 0

Answer:

Ammonium chloride

Explanation:

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A flask that weighs 450 g is filled with 145 ml of benzene. The weight of the flask and benzene is found to be 754 g. From this

vaieri [72.5K]

Answer:

Density, $d=2.09\ g/cm^3$

Explanation:

Given that,

Mass of a flask is 450 g

Volume of benzene added to the flask is 145 mL or 145 cm³

The weight of the flask and benzene is found to be 754 g.

We need to find the density of the benzene.

Weight of benzene added = total weight of flask and benzene-mass of flask

m = 754 g - 450 g

m = 304 g

Density = mass/volume

So,

$d=\dfrac{304\ g}{145\ cm^3}\\\\d=2.09\ g/cm^3$

So, the density of the benzene is $2.09\ g/cm^3$ .

7 0

3 years ago

Given the speed of light as 3.0 × 108 m/s, calculate the wavelength of the electromagnetic

kupik [55]

Answer:

The answer is

$\huge 4.054 \times {10}^{ - 5} \: m \\$

Explanation:

The wavelength of the electromagnetic

radiation can be found by using the formula

$\lambda = \frac{c}{f} \\$

f is the frequency

c is the speed of light

From the question we have

$\lambda = \frac{3.0 \times {10}^{8} }{7.4 \times {10}^{12} } \\$

We have the final answer as

$4.054 \times {10}^{ - 5} \: m$

Hope this helps you

4 0

3 years ago

A student heats a sample of hydrate once, and the mass of the sample and the evaporating dish is 16.428 g. After a second heatin

jolli1 [7]

Answer:

12.371 g

Explanation:

Given :

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g$

$m_{evaporating\ dish}+m_{First\ heated\ sample}=16.428\ g$

$m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g$

Mass of salt hydrate:

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Hydrate\ sample}=25.637\ g$

$m_{Hydrate\ sample}=25.637-m_{evaporating\ dish}\ g=25.637-1.135\ g=24.502\ g$

Mass of salt anhydrous:

$m_{evaporating\ dish}=1.135\ g$

$m_{evaporating\ dish}+m_{Second\ heated\ sample}=13.266\ g$

$m_{Second\ heated\ sample}=m_{salt\ anhydrous}=13.266-m_{evaporating\ dish}\ g=13.266-1.135\ g=12.131\ g$

Mass of water:

$m_{water}=m_{Hydrate\ sample}-m_{salt\ anhydrous}=24.502-12.131\ g=12.371\ g$

$m_{water}=12.371\ g$

4 0

4 years ago

A pressure of 125,400 pa is equal to what kPa

Goryan [66]

Answer:

125.4 kilopascals

Explanation:

wkdje

6 0

3 years ago

Read 2 more answers

How many grams of sodium chloride must dissolve in 750.0 g of water to make a 0.50 molal solution?

AleksandrR [38]

Molality is defined as the number of moles of solute in 1 kg of solvent.
molality of solution to be prepared is 0.50 molal
this means that in 1000 g of water there should be 0.50 mol of NaCl
if 1000 g of water should contain - 0.50 mol
then 750.0 g of water requires - 0.50 mol/kg x 0.750 kg = 0.375 mol
mass of NaCl in 0.375 mol - 58.5 g/mol x 0.375 mol = 21.9 g
therefore a mass of 21.9 g of NaCl is required

7 0

3 years ago