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2 years ago

When a surface is experiencing friction with another surface, how are the particles

Chemistry

1 answer:

Free_Kalibri [48]2 years ago

6 0

Answer:

the surface may become etheir scratchy and soft and that the surface may also be come heated up due to friction

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What geologic features might form in convergent boundaries Pls help

lisov135 [29]

1) deep trenches
2) active volcanoes
3) arcs of islands along the tectonic boundaries
4) in some cases, mountain ranges

5 0

3 years ago

The boiling point of water is 100.00 °C at 1 atmosphere.

Oksana_A [137]

Answer:

The solution's boiling point is 100,2°C and its molality is 0,13 m

Explanation:

This is the colligative propertie about elevation of boiling point

ΔT = Kb . m . i where

ΔT is the difference between T° at boiling point of the solution - T° at boiling point of the solvent pure

Kb means ebulloscopic constant (0,52 °C.kg/m .- a known value for water)

m means molality (moles of solute in 1kg of solvent)

i means theVan 't Hoff factor ( degree of dissociation for a compound)

IT HAS NO UNITS

NiI2 ---> Ni2+ + 2I- (we have 1 Ni2+ and 2 I-), the i for this, is 3

The 11,11 g of the salt are in 272,2g of water but I need to know how many mass of the salt is in 1000 g of water (1000 g is 1 kg) so the rule of three is:

272,2g ____ 11,11g

1000g _____ (1000 g . 11,11g) / 272,2g = 40,81g

As the molar mas of NiI2 is 312.5 g/mol, the moles of salt are, mass/molar mass, 40,81g /312.5 g/mol = 0,130 moles

T° of b p sl - 100°C = 0,52 °C.kg/m . 0,130 m/kg . 3

T° of boiling point solution = (0,52 °C.kg/m . 0,130 m/kg . 3) + 100°C

T° of boiling point solution = 100,2°C

4 0

3 years ago

2.17 in to 10^23 molecules of Br 2 to moles

KiRa [710]

I don't get what you are saying... Can you reword it?

3 0

2 years ago

Can someone help me with this please

dusya [7]

Boiling water: It starts bubbling and it becomes hot
Baking Cake: The cake becomes soft and not liquid
Burning Wood: The wood burns and turns into ash
Painting Door: The color changes and the wood numbs slightly

Hope this helps!
STSN

6 0

3 years ago

Read 2 more answers

Complete combustion of 7.40 g of a hydrocarbon produced 22.4 g of CO2 and 11.5 g of H2O. What is the empirical formula for the h

cluponka [151]

C2H5 First, you need to figure out the relative ratios of moles of carbon and hydrogen. You do this by first looking up the atomic weight of carbon, hydrogen, and oxygen. Then you use those atomic weights to calculate the molar masses of H2O and CO2. Carbon = 12.0107 Hydrogen = 1.00794 Oxygen = 15.999 Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488 Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087 Now using the calculated molar masses, determine how many moles of each product was generated. You do this by dividing the given mass by the molar mass. moles H2O = 11.5 g / 18.01488 g/mole = 0.638361 moles moles CO2 = 22.4 g / 44.0087 g/mole = 0.50899 moles The number of moles of carbon is the same as the number of moles of CO2 since there's just 1 carbon atom per CO2 molecule. Since there's 2 hydrogen atoms per molecule of H2O, you need to multiply the number of moles of H2O by 2 to get the number of moles of hydrogen. moles C = 0.50899 moles H = 0.638361 * 2 = 1.276722 We can double check our math by multiplying the calculated number of moles of carbon and hydrogen by their respective atomic weights and see if we get the original mass of the hydrocarbon. total mass = 0.50899 * 12.0107 + 1.276722 * 1.00794 = 7.400185 7.400185 is more than close enough to 7.40 given rounding errors, so the double check worked. Now to find the empirical formula we need to find a ratio of small integers that comes close to the ratio of moles of carbon and hydrogen. 0.50899 / 1.276722 = 0.398669 0.398669 is extremely close to 4/10, so let's reduce that ratio by dividing both top and bottom by 2 giving 2/5. Since the number of moles of carbon was on top, that ratio implies that the empirical formula for this unknown hydrocarbon is C2H5

3 0

3 years ago