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3 years ago

Which statement defines specific heat capacity for given sample?

Chemistry

2 answers:

ludmilkaskok [199]3 years ago

7 0

Answer:

D. the quantity of heat that is required to raise 1 g of the sample by 1*C (kelvin) at a constant pressure.

Explanation:

took the test

Aleks04 [339]3 years ago

6 0

Answer:

D. the quantity of heat that is required to raise 1 g of the sample by 1*C (kelvin) at a constant pressure.

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Which of the following is a combustion reaction?

azamat

Answer:

A.NaOH+HCI+NaCI+H2O

Explanation:

It's a Acid-base

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3 years ago

Conclusion Questions:

Lerok [7]

That’s the question?

5 0

3 years ago

If the freezing point of the solution had been incorrectly read 0.3 °C lower than the true freezing point, would the calculated

Dovator [93]

Answer : The molar mass of the solute would be low.

Explanation :

Formula used for depression in freezing point is:

$\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{w_b}{M_b}\times w_a}$

where,

$\Delta T_f$ = change in freezing point

$\Delta T_s$ = freezing point of solution

$\Delta T^o$ = freezing point of water

i = Van't Hoff factor

$K_f$ = freezing point constant

m = molality

$w_b$ = mass of solute

$w_a$ = mass of solvent

$M_b$ = molar mass of solute

From the formula we conclude that, when the freezing point of the solution read incorrectly that is freezing point of the solution is lower than the true freezing point then this means that change in freezing point would be high and the molar mass of the solute would be low.

Hence, the molar mass of the solute would be low.

6 0

4 years ago

02:22:18

Artemon [7]

Answer: b design procedure

Explanation:

design a procedure

7 0

3 years ago

Read 2 more answers

The reactant concentration in a second-order reaction was 0.560 m after 115 s and 3.30×10−2 m after 735 s . what is the rate con

Aloiza [94]

Use the formula for second order reaction:

$\frac{1}{C} = \frac{1}{C_0} + kt$

C = concentration at time t
C0 = initial conc.
k = rate constant
t = time

1st equation : $\frac{1}{0.56} = \frac{1}{C_0} + 115k$

2nd Equation: $\frac{1}{0.033} = \frac{1}{C_0} + 735k$

Find $\frac{1}{C_0}$ from 1st equation and put it in 2nd equation:

$\frac{1}{0.033} = \frac{1}{0.56} - 115k + 735k$

$\frac{1}{0.033} - \frac{1}{0.56} = 620k$

k = 0.046

3 0

3 years ago