A rock's mass is a:

Calculate the mass percent of H in C3H8O3. %

ki77a [65]

Answer:

because of it being that you have inferno dragon plsss

6 0

3 years ago

at what temperature will you find more potassium nitrate(kno3) in 100 g of H2O . In other words at which temperature will you ha

Oksana_A [137]

The answer to this question would be: 85c
Solubility is influenced by the temperature of the molecule. Higher temperature will result in a higher solubility. That is why brewing using hot water will be easier than using cold water. Higher temperature also means higher reaction rate.

4 0

3 years ago

Hydrogen peroxide can be prepared by the reaction of barium peroxide with sulfuric acid according to the reaction BaO2(s)+H2SO4(

Vesna [10]

Answer:

VH2SO4 = 145.3 mL

Explanation:

Mw BaO2 = 169.33 g/mol

⇒ mol BaO2 = 53.5g * ( mol BaO2 / 169.33 g BaO2) = 0.545 mol BaO2

⇒according to the reaction:

mol BaO2 = mol H2SO4 = 0.545 mol

⇒ V H2SO4 = 0.545 mol H2SO4 * ( L H2SO4 / 3.75 mol H2SO4 )

⇒V H2SO4 = 0.1453 L (145.3 mL)

6 0

3 years ago

Read 2 more answers

A rectangular block has the following dimensions: 3.21 dm, 5.83 cm, and 1.84 in. The block has a mass of 1.94 kg. What is the de

spin [16.1K]

The density of the rectangular block in g/mL is 7.0.

Given the following data:

Mass of block = 22.8 gra1.94 kg

Length of block = 3.21 cm

Width of block = 5.83 cm

Height of block = 1.84 in.

To find the density of the block in g/mL:

First of all, we would determine the volume of the rectangular block by using the following formula:

$Volume = length$ × $width$ × $height$

Conversion:

1 in = 2.54 cm

5.83 in = X cm

Cross-multiplying, we have:

$X = 2.54(5.83)\\\\X = 14.81 \; cm$

$Volume = 3.21$ × $5.83$ × $14.81$

Volume = 277.16 cubic centimeters.

Note: Milliliter (mL) is the same as cubic centimeters.

1000 grams = 1 kg

Y grams = 1.94 kg

Cross-multiplying, we have:

Y = 1940 grams

Now, we can find the density:

$Density = \frac{Mass}{Volume}\\\\Density = \frac{1940}{277.16}$

Density = 7.0 g/mL

Therefore, the density of the rectangular block in g/mL is 7.0.

4 0

3 years ago

How many moles of sodium carbonate are contained by 57.3g of sodium carbonate

Lady_Fox [76]

Answer:

$\boxed {\boxed {\sf 0.541 \ mol \ Na_2CO_3}}$

Explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.

We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.

Na: 22.9897693 g/mol
C: 12.011 g/mol
O: 15.999 g/mol

Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

Na₂ = 22.9897693 * 2= 45.9795386 g/mol
O₃ = 15.999 * 3= 47.997 g/mol
Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol

We will convert using dimensional analysis. Set up a ratio using the molar mass.

$\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}$

We are converting 57.3 grams to moles, so we multiply by this value.

$57.3 \ g \ Na_2CO_3} *\frac {105.9875386 \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}$

Flip the ratio so the units of grams of sodium carbonate cancel.

$57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386 \ g \ Na_2CO_3}$

$57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }$

$\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3$

$0.5406295944 \ mol \ Na_2CO_3$

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

$0.541 \ mol \ Na_2CO_3$

There are approximately 0.541 moles of sodium carbonate in 57.3 grams.

6 0

2 years ago