How much so you can be dissolved and 50 mL of water at 30 Celsius see chart:)

Answer:

troposphere, stratosphere, mesosphere, thermosphere and exosphere

Explanation:

Answer:

23.8g of sodium phosphate are formed

Explanation:

Based on the reaction of sodium, Na, with phosphoric acid, H₃PO₄:

3Na + H₃PO₄ → Na₃PO₄ + 3/2 H₂

3 moles of sodium produce 1 mole of sodium phosphate

To solve this question we must find the moles of sodium in 10g. Using the chemical reaction we can find the moles -And the mass- of sodium phosphate produced, as follows:

Moles Na -Molar mass: 22.99g/mol-

10g * (1mol / 22.99g) = 0.435 moles Na

Moles Na₃PO₄:

0.435 moles Na * (1mol Na₃PO₄ / 3mol Na) = 0.145 moles Na₃PO₄

Mass Na₃PO₄ -Molar mass: 163.94g/mol-

0.145 moles Na₃PO₄ * (163.94g/mol) =

Answer:

The mass of the solution is 140 grams.

Explanation:

The mass of the solution is given by the sum of the solute's mass and the solvent's mass:

[tex] m_{sol} = m_{ste} + m_{svt} [/tex]

Where:

[tex] m_{sol} [/tex]: is the mass of the solution

[tex]m_{ste}[/tex]: is the mass of the solute

[tex]m_{svt} [/tex]: is the mass of the solvent

The solute is the salt (40.0 g) and the solvent is the water (100 g), so:

[tex] m_{sol} = m_{ste} + m_{svt} = 40 g + 100 g = 140 g [/tex]

Therefore, the mass of the solution is 140 grams.

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Answer:

2.7 atm = 2052 torr

I hope this helps you.

Answer:

2052 torr

Explanation:

Multiply 2.7 atm by the conversion factor to torr which is 760 torr = 1 atm

Answer:

1.238 years will it take

Explanation:

The isotope decay of an atom follows the equation:

Ln[A] = -kt + ln[A]₀

Where [A] could be taken as the amount of the isotope after t years

[A] = 1.00mg

t = ?

k is the decay constant; k = ln2 / Half-life

k = ln 2 / 2.86 years = 0.2424 years⁻¹

[A]₀ is the initial amount of the isotope = 1.35mg

Replacing:

ln1 = -0.2424years⁻¹*t + ln 1.35

0.2424years⁻¹*t = ln 1.35

t =

Explanation:

If the density of the object is more than that of water, it will sink. Otherwise it will float. The density of water is 1 g/mL.

Substance 1,

Mass, m = 450 g, Volume, V = 90 mL

Density = mass/volume

So,

[tex]d_1=\dfrac{450}{90}\\\\=5\ g/mL[/tex]

It will sink.

Substance 2,

Mass, m = 35 g, Volume, V = 70 mL

Density = mass/volume

So,

[tex]d_2=\dfrac{35}{70}\\\\=0.5\ g/mL[/tex]

It will float.

Substance 3,

Mass, m = 24 g, Volume, V = 12 mL

Density = mass/volume

So,

[tex]d_3=\dfrac{24}{12}\\\\=2\ g/mL[/tex]

It will sink.

Answer:

50.63 cm³

Explanation:

Using Charles law equation as follows:

V1/T1 = V2/T2

Where;

V1 = initial volume (cm³)

V2 = final volume (cm³)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the question, V1 = 45.0 cm³, V2 = ?, T1 = 280 K, T2 = 315 K

Using; V1/T1 = V2/T2

45/280 = V2/315

Cross multiply

45 × 315 = 280 × V2

14175 = 280V2

V2 = 14175 ÷ 280

V2 = 50.625

V2 = 50.63 cm³

Answer:

51,069.2 J

Explanation:

Use the heat of vaporization formula:

q=m⋅Hv

The heat of vaporization of ethanol is 854 J/g (given in the question), which means that it takes 854 joules of energy to vaporize one gram of ethanol.

Since we have 59.8 grams of ethanol, plug 59.8 into the formula and multiply by 854.

q = (59.8) (854)

q= 51069.2 J

hope this helped somehow

Answer:

Explanation:

The answer is 202

Answer:

the 6 stands for the atomic number

the 12.01 stands for the atomic mass

Answer: At a temperature of 34.44 K the volume and pressure of this gas become one-third of their initial values.

Explanation:

Given : [tex]V_{1}[/tex] = 1.2 L,        [tex]T_{1} = 37^{o}C = (37 + 273) K = 310 K[/tex]

[tex]P_{1}[/tex] = 3 atm,                  [tex]V_{2} = \frac{1.2}{3} = 0.4 L[/tex] ,      [tex]P_{2} = \frac{3}{3} = 1 atm[/tex]

Formula used to calculate the final temperature is as follows.

[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]

Substitute the values into above formula as follows.

[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{3 atm \times 1.2 L}{310 K} = \frac{1 atm \times 0.4 L}{T_{2}}\\T_{2} = \frac{1 atm \times 0.4 L \times 310 K}{3 atm \times 1.2 L}\\= \frac{124}{3.6} K\\= 34.44 K[/tex]

Thus, we can conclude that the final temperature is 34.44 K.

Answer:

IM just too pogchamp

Explanation:

Answer: Carbon-14 decays quickly leaving the amount too small to measure.

Explanation:

Answer:

b

Explanation:

i did it

Answer:

3 moles of calcium oxide are formed

Explanation:

Based on the reaction of Calcium Ca, with oxygen, O₂, to produce calcium oxide, CaO is:

2Ca + O₂ → 2CaO

where 2 moles of calcium produce 2 moles of Calcium oxide.

That means, if 3 moles of Ca react in presence of oxygen:

3 moles of calcium oxide are formed

Answer:

Red supergiant

Explanation:

Their core is the hottest.

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Answer:

3.840 atm

Explanation:

We use Boyle's law, which relates the pressure (P) and the volume (V) of a gas at a constant temperature. The change from initial P and V (P₁ and V₁) to final P and V (P₂ and V₂) is expressed as:

P₁V₁ = P₂V₂

We have the following data:

P₁= 0.960 atm

V₁= 200.0 mL

V₂ = 50.0 mL

Thus, we introduce the data in the equation and calculate the final pressure P₂:

P₂ = P₁V₁/V₂= (0.960 atm x 200.0 mL)/50.0 mL = 3.840 atm

Therefore, a pressure of 3.840 atm is needed to reduce the volume of the gas from 200.0 mL to 50.0 mL.

Answer:

Hydrogen (since this is all the info I'm working with)

Explanation:

Since we need to know how which substance has the most amount of molecules, we would need to look at the Atomic masses of these elements. Hydrogen has an atomic mass of 1.01g/mol whereas something like oxygen has an atomic mass of 16.00g/mol. Judging this, we can determine since a mole is made up of 6.02*10^23 molecules(Avogadro's number), and we have 1/1.01grams of hydrogen, we can deduce that Hydrogen would contain the largest amount of molecules

Answer:

29.42 Litres

Explanation:

The general/ideal gas equation is used to solve this question as follows:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K

According to the information provided in this question;

mass of nitrogen gas (N2) = 25g

Pressure = 0.785 atm

Temperature = 315K

Volume = ?

To calculate the number of moles (n) of N2, we use:

mole = mass/molar mass

Molar mass of N2 = 14(2) = 28g/mol

mole = 25/28

mole = 0.893mol

Using PV = nRT

V = nRT/P

V = (0.893 × 0.0821 × 315) ÷ 0.785

V = 23.09 ÷ 0.785

V = 29.42 Litres

Answer:

Graphene is made of pure carbon.

Explanation:

It is a material in which the carbon atoms are arranged in a single layer, creating a honeycomb pattern. It should be stressed that this layer of carbon is only one atom thick, although some authors consider up to ten layers of carbon to be graphene.

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Answer:

The mass of oxygen will be 19.53g

Answer:

B

Explanation:

four electron in atom in cal

Answer:

welding torch fuel  

Explanation:

Alkenes are chemical compounds formed by a carbon atom and a hydrogen atom. They have an open chain with double bonds between two carbon acts. They are widely used in the composition of fuels, such as welding torch fuel and have as main characteristics the solubility in solvents, the insolubility in water, high melting and boiling point and the absence of colors.

Welding torch fuel is composed of alkenes. The alkenes comprise a series of compounds that are composed of carbon and hydrogen atoms with at least one double bond in the carbon chain.

Alkenes are a homologous series of hydrocarbons that contain a carbon-carbon double bond.

The number of hydrogen atoms in an alkene is double the number of carbon atoms, so they have the general formula [tex]C_nH _2 _n[/tex].

For example, the molecular formula of ethene is [tex]C _2 H _4[/tex] , while for propene it is [tex]C _3 H _6[/tex].

Alkenes are chemical compounds formed by a carbon atom and a hydrogen atom.

They have an open chain with double bonds between two carbon acts.

They are widely used in the composition of fuels, such as welding torch fuel and have as main characteristics the solubility in solvents, the insolubility in water, high melting and boiling point and the absence of colors.

Hence, the Welding torch fuel is composed of alkenes.

Learn more about the alkenes here:

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Answer:

As nouns the difference between impression and imprint

is that impression is the indentation or depression made by the pressure of one object on or into another while imprint is an impression; the mark left behind by printing something.

The term imprint is defined as to mark by or as if by pressure and the term impression is defined as the effect produced by impressing.

The difference between an imprint and an impression is that the imprint is the mark left behind when anything is printed, whereas the impression is the indentation or depression caused by the pressure of one object against or into another.

While a shoe won't make an impression in loose sand, it will on a hard surface like concrete or linoleum. These imprints or impressions might be contrasted with the shoe of a suspect or the tire of a car to see if they belong to the same object.

Thus, The term imprint is defined as to mark by or as if by pressure and the term impression is defined as the effect produced by impressing.

To learn more about the imprint and the impression, follow the link;

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Answer:

Explanation:

If you would like to solve 1/(x + 3) = (x + 10)/(x - 2), you can do this using the following steps:

1/(x + 3) = (x + 10)/(x - 2)

1 * (x - 2) = (x + 10) * (x + 3)

x - 2 = x^2 + 3x + 10x + 30

0 = x^2 + 13x + 30 - x + 2

0 =  x^2 + 12x + 32

0 = (x + 8) * (x + 4)

1. x = -8

2. x = -4

Respuesta:

90.0 %

Explicación:

Paso 1: Escribir la ecuación química balanceada

N₂ + 3 H₂ ⇒ 2 NH₃

Paso 2: Calcular el rendimiento teórico de NH₃ a partir de 140 g de N₂

En la ecuación balanceada, participan de N₂: 1 mol × 28.01 g/mol = 28.01 g y de NH₃: 2 mol × 17.03 g/mol = 34.06 g.

140 g N₂ × 34.06 g NH₃ /28.01 g N₂ = 170 g NH₃

Paso 3: Calcular el rendimiento porcentual de NH₃

El rendimiento experimental de NH₃ es 153 g. Podemos calcular el rendimiento porcentual usando la siguiente fórmula.

R% = rendimiento experimental / rendimiento teórico × 100%

R% = 153 g / 170 g × 100% = 90.0 %