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3 years ago

PLEASEEEEEEEE ASAPPPPPWhat is the density of a piece of cardboard that has a mass of 250 g and volume of 46 mL? *

Chemistry

2 answers:

FromTheMoon [43]3 years ago

8 0

Answer:

5.4347826087

Explanation:

The formula for density is Mass/Volume so you would do 250/46 to get the answer of 5.4347826087 grams per milliliter

trapecia [35]3 years ago

6 0

HEELP ME HELP ME HELP ME HELP ME HELP ME

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Strike441 [17]

Answer:

Scientist used models to explain and predict the behavior of real object or system

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2 years ago

What is the volume in liters of 321 g of liquid with a density of 0.84 g/mL

Rasek [7]

Density is weight by volume.

First. If you divide the weight by density you can find the volume

Second you must convert the ML in to Liters.

$\frac{321 \frac{g}{1}}{0.84 \frac{g}{mL}} = \frac{321(g)(mL)}{0.84g}=\frac{321mL}{0.84}=382.14mL$

1L=1000ml

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

$(382.14mL)( \frac{1L}{1000mL})= \frac{(382.14mL)(L)}{1000mL} =\frac{382.14L}{1000}=0.38214L$

0.38214 Liters.

4 0

3 years ago

How many grams of copper (II) nitrate would be produced from 0.80 g of copper metal reacting with excess nitric acid?

zaharov [31]

Answer:

$m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2$

Explanation:

Hello!

In this case, since the chemical reaction between copper and nitric acid is:

$2HNO_3+Cu\rightarrow Cu(NO_3)_2+H_2$

By starting with 0.80 g of copper metal (molar mass = 63.54 g/mol) and considering the 1:1 mole ratio between copper and copper (II) nitrate (molar mass = 187.56 g/mol) we can compute that mass via stoichiometry as shown below:

$m_{Cu(NO_3)_2}=0.80gCu*\frac{1molCu}{63.54gCu} *\frac{1molCu(NO_3)_2}{1molCu} *\frac{187.56gCu(NO_3)_2}{1molCu(NO_3)_2} \\\\m_{Cu(NO_3)_2}=2.36 gCu(NO_3)_2$

However, the real reaction between copper and nitric acid releases nitrogen oxide, yet it does not modify the calculations since the 1:1 mole ratio is still there:

$4HNO_3+Cu\rightarrow Cu(NO_3)_2+2H_2O+2NO_2$

Best regards!

7 0

3 years ago

When a mixture of 12.0 g of acetylene (c2h2 and 12.0 g of oxygen (o2 is ignited, the resultant combustion reaction produces co2

valina [46]

The product of the complete combustion of any fuel (in this case, acetylene) are indeed water and carbon dioxide.
Balancing the combustion reaction,
C2H2 +(5/2) O2 --> 2CO2 + H2O
The number of moles of C2H2 will be,
(12 g) x (1 mole/26 g) = 6/13 mole
Then, the number of moles of O2 is,
(12 g) x (1 mole/32 g) = 3/8 mole
Therefore the limiting reaction is the O2. Getting the amount of CO2 and H2O produced from balancing,
CO2 = (3/8 moles) x (2 moles CO2/ 5/2 mole O2)(44 g/ 1 mole) = 52.8 g
H2O = (3/8 moles) x (1 mole / 5/2 mole O2)(18 g / 1 mole) = 2.7 g

3 0

3 years ago

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Answer:

The anwer is not D the anwer is A

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