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djverab [1.8K]
2 years ago
15

How do you find the mass of a tissue box? NOT the volume. I need this for an assignment. please help!

Chemistry
1 answer:
Alina [70]2 years ago
8 0
  • First you need to find the volume
  • Make sure Density of the box is given .

  • Now Use density formula

\\ \sf\longmapsto Density=\dfrac{Mass}{Volume}

\\ \sf\longmapsto Mass=Density\times Volume

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Herbicides break down dead organisms into humus. True or False?
Rzqust [24]

Answer:

true

Explanation:

8 0
2 years ago
HELPP ME PLEASE!! What would be the density of a 40 g object that displaces 10 ml of water?
Snezhnost [94]

Answer:

<h2>The answer is 4.0 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 40 g

volume = 10 mL

The density of the object is

density =  \frac{40}{10}  = 4 \\

We have the final answer as

<h3>4.0 g/mL</h3>

Hope this helps you

5 0
3 years ago
How many levels will argon have?
Varvara68 [4.7K]
They have about 4 or 6
7 0
3 years ago
A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio
Pavlova-9 [17]

Answer:

C_7H_6O_2

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC

And the moles of hydrogen due to the produced 1.50 grams of water:

n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}  =0.166molH

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO

Thus, the subscripts in the empirical formula are:

C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol

Which are equal to the moles of the acid:

n_{acid}=0.0279mol

And the molar mass:

MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

C_7H_6O_2

Best regards!

8 0
3 years ago
Student perfotms a Benedict's test on an unknown substance. He adds reagent(the chemical required to make a color change), and n
elixir [45]

Answer:

Reducing sugars are absent

Explanation:

Benedict's solution is an substance used in testing sugars. It is mixture of sodium carbonate, sodium citrate and copper(II) sulfate pentahydrate. It can be used instead of Fehling's solution in testing for the presence of reducing sugars.

Reducing sugars contain the -CHO group. If there is no colour change after the addition of Benedict's solution, then we can conclude that reducing sugars are absent.

7 0
3 years ago
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