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3 years ago

If told to determine whether or an unknown substance is either salt or sugar what do you do to determine which it is explain?

1 answer:

3 0

If you were allowed to taste it sugar will be more sweeter whereas salt wouldnt. But if you were to touch it then sugar is more tinnier than salt.

You might be interested in

calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w

kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is $-22Jmole^-^1k^-^1$

(b) The normal boiling point of water (J·K−1·mol−1) is $-109Jmole^-^1K^-^1$

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is 109J/mole

Explanation:

Lets calculate

(a) - General equation -

$(\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p$ = $-5_m(\beta )+5_m(\alpha )$ = $-\frac{\Delta H}{T}$

$\alpha ,\beta$ → phases

ΔH → enthalpy of transition

T → temperature transition

$(\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p$ = $= -\frac{\Delta_fH}{T_f}$

= $\frac{-6.008kJ/mole}{273.15K}$ ( $\Delta_fH$ is the enthalpy of fusion of water)

= $-22Jmole^-^1k^-^1$

(b) $(\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}$

= $\frac{40.656kJ/mole}{373.15K}$ ( $\Delta_v_a_p_o_u_rH$ is the enthalpy of vaporization)

= $-109Jmole^-^1K^-^1$

$[\mu(l-5$ ° $C)-\mu(l,0$ ° $C)]$ = $[\mu(s-5$ ° $C)-\mu(s,0$ ° $C)]$ $=-S_m$ ΔT

$\mu(l,-5$ ° $C)-\mu(s,-5$ ° $C)=-Sm\DeltaT [\mu(l,0$

$\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)$

= 109J/mole

6 0

2 years ago

A sample of oxalic acid (a diprotic acid of the formula H2C2O4) is dissolved in enough water to make 1.00 L of solution. A 100.0

OleMash [197]

Answer: The mass of original oxalic acid sample is 6.75 grams

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2C_2O_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=?M\\V_1=100.0mL\\n_2=1\\M_2=0.750M\\V_2=20.0mL$

Putting values in above equation, we get:

$2\times M_1\times 100.0=1\times 0.750\times 20.0\\\\M_1=\frac{1\times 0.750\times 20.0}{2\times 100.0}=0.075M$

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of oxalic acid = ? g

Molar mass of oxalic acid = 90 g/mol

Molarity of solution = 0.075 M

Volume of solution = 1.00 L

Putting values in above equation, we get:

$0.075M=\frac{\text{Mass of oxalic acid}}{90g/mol\times 1L}\\\\\text{Mass of oxalic acid}=(0.075\times 90\times 1)=6.75g$

Hence, the mass of original oxalic acid sample is 6.75 grams

7 0

3 years ago

The addition of an inert gas has no effect on the equilibrium position of a gaseous reaction because ______. Multiple choice que

boyakko [2]

Inert gas does not affect the equilibrium position:

It is because the partial pressures of the reaction components remain the same.

What is Inert Gas?

Under a given set of conditions, an inert gas is a gas that does not undergo chemical reactions.
The noble gases (helium, neon, argon, krypton, xenon, and radon) were previously known as "inert gases" due to their perceived lack of involvement in any biochemical processes.
Because inert gases are non-reactive, they do not affect equilibrium partial pressures and thus do not affect volume.
An inert gas does not react with the reactants or products; it does not change the concentration of the products and reactants. Furthermore, because the volume is constant, the concentrations are unaffected. As a result, this does not affect equilibrium.

The equilibrium position won't change if an inert gas is added. A volume change won't change the equilibrium position if the total moles of gas in the products and reactants are the same. When the volume is reduced, the process changes to create fewer moles of gas.

Learn more about the inert gas here,

brainly.com/question/15909389

#SPJ4

5 0

2 years ago

Question 10(Multiple Choice Worth 5 points)

jarptica [38.1K]

Answer:

O Generation of electricity

Explanation:

because the others dont make sense to me and tbh carbon is used everyway and everywhere

7 0

2 years ago

Using the volumes of sodium thiosulfate solution you just entered, the mass of bleach sample, and the average molarity of the so

dedylja [7]

Answer:

3.18 (w/w) %

Explanation:

In the problem, you can find mass of NaClO knowing the reaction of NaClO with Na₂S₂O₃ is:

NaClO + 2Na₂S₂O₃ + H₂O → NaCl + Na₂S₄O₆ +2NaOH + NaCl

Where 1 mole of NaClO reacts with 2 moles of Na₂S₂O₃

Moles of thiosulfate in the titration are:

0.0101L ₓ (0.042mol / L) = 4.242x10⁻⁴ moles of Na₂S₂O₃

Thus, moles of NaClO in the initial solution are:

4.242x10⁻⁴ moles of Na₂S₂O₃ ₓ (1mol NaClO / 2 mol Na₂S₂O₃) = 2.121x10⁻⁴ moles NaClO

As molar mass of NaClO is 74.44g/mol, mass of 2.121x10⁻⁴ moles are:

2.121x10⁻⁴ moles ₓ (74.44g / mol) = 0.0158g of NaClO

As mass of bleach is 0.496g, mass percent is:

0.0158g NaClO / 0.496g bleach ₓ 100 =

4 0

3 years ago