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2 years ago

if the mass of a box was 400 grams, and the length is 2cm the width is 2cm and the height is 1cm. what is the density?

Chemistry

1 answer:

oksano4ka [1.4K]2 years ago

4 0

Answer:

The density of the box is 6.25g/cm³

Explanation:

400 ÷ 4³ = 6.25

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The molarity of a solution containing 7.1 g of sodium sulfate in 100 mL of an aqueous solution is

Sergio [31]

Answer:

0.5 M

Explanation:

First we have to start with the molarity equation:

$M=\frac{mol}{L}$

We need to know the amount of moles and the litters.

If we have 100 mL we can convert this value to “L”, so:

$100~mL\frac{1~L}{1000~mL}$

$0.1~ L$

Now we can continue with the moles, for this we have to know the formula of sodium sulfate $Na_2SO_4$ , with this formula we can calculate the molar mass if we know the atomic mass of each atom on the formula (Na: 23 g/mol, S: 32 g/mol, O: 16 g/mol). We have to multiply each atomic mass by the amount of atoms in the formula, so:

$molar~ mass~=~ (23*2)+(32*1)+(16*4)= ~ 142~ g/mol$

In other words:

$1~mol~ Na_2SO_4=~142~g~ of~Na_2SO_4$

Now we can calculate the moles:

$7.1~g~ of~Na_2SO_4\frac{1~mol~ Na_2SO_4}{142~g~ of~Na_2SO_4}$

$0.05~mol~ Na_2SO_4$

Finally, we can calculate the molarity:

$M=\frac{0.05~mol~ Na_2SO_4 }{0.1~ L}$

$M=0.5$

I hope it helps!

4 0

2 years ago

Read 2 more answers

What are the kingdoms of eukaryotes

Talja [164]

Protists,fungi,plantar and animalia

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2 years ago

a sample of a compound contains 60gC and 5.05gH.it molar mass is 78.12g/mol what is the compound molecular formula

elena55 [62]

(60)/(60+5.05)=.922367 C
1-0.922367=0.07763259 H
(0.922367)(78.12)=72.05534204 C
(0.07763259)(78.12)=6.06 H
72.05534204/(12.01)=6 C
6.06/1.01=6 H
Empirical= CH
Molecular=C6H6

5 0

2 years ago

Wood and a cement block are considered to be transparent and are good conductors of electricity.

Tanzania [10]

Answer:

ITS FALSE electricity doesn't travel through wood same with cement!

3 0

2 years ago

What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the su

anastassius [24]

Answer:

$V=27992L=28.00m^3$

Explanation:

Hello,

In this case, the combustion of methane is shown below:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

$n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4$

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

$PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3$

Best regards.

8 0

2 years ago