Are poppy seeds an element, compound, or mixture?

What particle(s) give an atom its mass number?

FinnZ [79.3K]

The Higgs Boson is the only known particle that gives mass to other particles/atoms. Particle physicists and theoretical physicists are arguing that there are more than one type of Higgs Boson.

6 0

3 years ago

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When she placed her hand on the side of the container, it felt really cold, but it didn't feel that way before. The temperature

artcher [175]

Answer:

this is a physical reaction

Explanation:

3 0

3 years ago

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Calculate the values of ΔU, ΔH, and ΔS for the following process:

kykrilka [37]

The values of the changes are

ΔH = 46.44kJ

Δu = 43.34Kj

ΔS = 126.6 J/K

<h3>How to find change in H</h3>

= 1 mol + 75.3 (100 + 273) - 25 + 273 + 1 * 40.79

= 5.6475 + 40.79

= 46.44kJ

<h3>How to find change in S</h3>

1 mol + 75.3 x ln 373/298 + 1 mol x 40.79

= 0.1263

ΔS = 126.6 J/K

<h3>How to find the change in U</h3>

46.44 - 0.00831 * 373

= 43.34Kj

Read more on molar heat here:

brainly.com/question/13439286

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5 0

2 years ago

A compound that contains only carbon, hydrogen, and oxygen is 58.8% C and 9.87% H by mass. What is the empirical formula of this

Dmitry [639]

<u>Answer:</u> The empirical formula of the compound becomes $C_5H_{10}O_2$

<u>Explanation:</u>

The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.

Let the mass of the compound be 100 g

Given values:

% of C = 58.8%

% of H = 9.87%

% of O = [100 - 58.8 - 9.87] = 31.33%

Mass of C = 58.8 g

Mass of H = 9.87 g

Mass of O = 31.33 g

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

To formulate the empirical formula, we need to follow some steps:

<u>Step 1:</u> Converting the given masses into moles.

Molar mass of C = 12 g/mol

Molar mass of H = 1 g/mol

Molar mass of O = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of C}=\frac{58.8g}{12g/mol}=4.9 mol$

$\text{Moles of H}=\frac{9.87g}{1g/mol}=9.87 mol$

$\text{Moles of O}=\frac{31.33g}{16g/mol}=1.96mol$

<u>Step 2:</u> Calculating the mole ratio of the given elements.

Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 1.96 moles

$\text{Mole fraction of C}=\frac{4.9}{1.96}=2.5$

$\text{Mole fraction of H}=\frac{9.87}{1.96}=5.03\approx 5$

$\text{Mole fraction of O}=\frac{1.96}{1.96}=1$

Converting the mole fraction into whole numbers by multiplying them with 2.

$\text{Mole fraction of C}=2.5\times 2=5$

$\text{Mole fraction of H}=5\times 2=10$

$\text{Mole fraction of O}=1\times 2=2$

<u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 5 : 10 : 2

Hence, the empirical formula of the compound becomes $C_5H_{10}O_2$

8 0

3 years ago

Which aqueous solution would have the lowest vapor pressure at 25°c 1 M NaCl?

aleksley [76]

The question is incomplete, here is a complete question.

Which aqueous solution would have the lowest vapor pressure at 25°c.

A) 1 M NaCl

B) 1 M $K_3PO_4$

C) 1 M $C_{12}H_{10}O_{11}$

D) 1 M $MgCl_2$

E) 1 M $C_6H_{12}O_6$

Answer : The correct option is, (B) 1 M $K_3PO_4$

Explanation :

According to the relative lowering of vapor pressure, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component of the solution multiplied by the vapor pressure of that component in the pure state.

1 M means that the 1 moles of solute present in 1 liter of solution.

Formula used :

$\frac{\Delta p}{p^o}=i\times X_B$

where,

$p^o$ = vapor pressure of the pure component (water)

$p_s$ = vapor pressure of the solution

$X_B$ = mole fraction of solute

i = Van't Hoff factor

As we know that the vapor pressure depends on the mole fraction of solute and the Van't Hoff factor.

So, the greater the number of particles of solute dissolved the lower the resultant vapor pressure.

(a) The dissociation of 1.0 M $NaCl$ will be,

$NaCl\rightarrow Na^++Cl^-$

So, Van't Hoff factor = Number of solute particles = $Na^++Cl^-$ = 1 + 1 = 2

(b) The dissociation of 1 M $K_3PO_4$ will be,

$K_3PO_4\rightarrow 3K^{+}+PO_4^{3-}$

So, Van't Hoff factor = Number of solute particles = $3K^{+}+PO_4^{3-}$ = 3 + 1 = 4

(c) The dissociation of 1 M $C_{12}H_{10}O_{11}$ is not possible because it is a non-electrolyte solute. So, the Van't Hoff factor will be, 1.

(d) The dissociation of 1.0 M $MgCl_2$ will be,

$MgCl_2\rightarrow Mg^{2+}+2Cl^{-}$

So, Van't Hoff factor = Number of solute particles = $Mg^{2+}+2Cl^{-}$ = 1 + 2 = 3

(e) The dissociation of 1 M $C_6H_{12}O_{6}$ is not possible because it is a non-electrolyte solute. So, the Van't Hoff factor will be, 1.

From this we conclude that, 1 M $K_3PO_4$ has the highest Van't Hoff factor which means that the solution will exhibit the lowest vapor pressure.

Hence, the correct option is, (B) 1 M $K_3PO_4$

8 0

3 years ago