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3 years ago

Compare the charges and masses of protons neutrons and electrons

1 answer:

8 0

Charges is then there is a negative sign at the right corner of an element it means you have to add, and when it is positive, it means you have to substract

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2.00 L of 0.500 M H2SO4 (density = 1.030 g/mL) at 25.000°C is mixed with 200. g of a NaOH solution, also at 25.000°C. After stir

AfilCa [17]

Answer:

Q = 111.8 kJ;

The reaction is endothermic.

Explanation:

The heat of the mixture can be calculated by the equation:

Q = m*c*ΔT

Where Q is the heat, m is the mass of the solution, c is the specific heat, and ΔT is the variation in the temperature (fina - initial).

The mass of H₂SO₄ is the volume multiplied by its density:

mH₂SO₄ = 2000 mL * 1.030 g/mL = 2060 g

Then the total mass is:

m = 2060 + 200

m = 2260 g

Q = 2260*4.184*(36.820 - 25.000)

Q = 111,468.03 J

Q = 111.8 kJ

The reaction is endothermic because the solution is absorbing heat (Q > 0), which can be noticed by the increase in the temperature. When the reaction loses heat, it is exothermic and the temperature decreases.

4 0

3 years ago

The water for the pan is not saturated with co2. will the reported percent caco3 in the original sample be too high, too low, or

enot [183]

Co2 will dissolve in water if water is saturated with Co2 first then it will absorb some of CO2 it appear as less is absorb that was actually the case . this would make it appear as present of CACO3 is in original sample is lower than it will be too low

8 0

4 years ago

What is the density of an object if its mass is 10 g and a volume of 2 mL?

Lilit [14]

Hi there! Let's solve this problem shall we!

⠀Volume = 10g

Mass = 2 mL

In this specific problem, they are asking us to find the density of the object. So, using the information given to us (volume and mass), let's solve the problem!

Now, if you remember, D = M ÷ V

So, let's fill in the blanks!

D = Our unknown value

M = 2mL

V = 10g

Here is the filled out formula:

D = M ÷ V

D = 2mL ÷ 10g

D = 5 g/mL

*Make sure you put the units for your final solution!*

4 0

2 years ago

Calculate the solubility of benzene in water at 25 c in ppm. the required henry's law constant is 5.6 bar/mol/kg and benzene's s

kap26 [50]

The relationship between pressure and solubility of the gas is given by Henry's law as:

$S_g = kP_g$

where,

$S_g$ is the solubility of the gas.

$k$ is proportionality constant i.e. Henry's constant.

$P_g$ is pressure of the gas.

$k = 5.6 bar/mol/kg$ (given)

$P_g = 0.13 bar$ (given)

Substituting the values,

$S_g = 5.6 bar/mol/kg\times 0.13 bar = 0.728 mole/kg$

To convert $mole/kg$ to $g/kg$ :

Molar mass of benzene, $C_6H_6$ = $6\times 12+6\times 1 = 78 g/mol$

$0.728\times 78 = 56.784 g/kg$

Now for converting into $ppm$ :

Since, $1 ppm = 0.001 g/kg$

So, $56.784\times 1000 = 56784 ppm$ .

Hence, the solubility of benzene in water at $25^{o} C$ in $ppm$ is $56784 ppm$ .

7 0

3 years ago

Enter ionic and net equations: feso4(aq)+ na3po4(aq) arrow fe3(po4)2(s)+na2so4(aq)

stepan [7]

Answer:

ionic equation : 3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)

net ionic equation: 3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)

Explanation:

The balanced equation is

3FeSO4(aq)+ 2Na3PO4(aq) → Fe3(PO4)2(s)+ 3Na2SO4(aq)

Ionic equations: Start with a balanced molecular equation. Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions . Indicate the correct formula and charge of each ion. Indicate the correct number of each ion . Write (aq) after each ion .Bring down all compounds with (s), (l), or (g) unchanged. The coefficents are given by the number of moles in the original equation

3Fe(2+)(aq) + 3SO4(2-)(aq)+ 6Na(+)(aq) + 2PO4 (3-) (aq) → Fe3(PO4)2(s)+ 6Na(+) + 3SO4(2-)(aq)

Net ionic equations: Write the balanced molecular equation. Write the balanced complete ionic equation. Cross out the spectator ions, it means the repeated ions that are present. Write the "leftovers" as the net ionic equation.

3Fe(2+)(aq) + 2PO4 (3-)(aq) → Fe3(PO4)2(s)

6 0

3 years ago