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snow_tiger [21]

2 years ago

12

3. If a gas has a pressure of 32.1 psi at a temperature of 25°C, then what is the new pressure if the temperature is increased t

o 75°C?

1 answer:

mihalych1998 [28]2 years ago

6 0

Answer:

baho kag bilat sige ka ug palubut

Explanation:

sige ka ug eyut

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J.J. Thomson's plum pudding model of the atom followed the discovery of

Rama09 [41]

The answer is A i think from the options though the answers are worded a bit weird

3 0

3 years ago

Read 2 more answers

10) The presence of nitrates in soil can be shown by warming the soil with

tresset_1 [31]

Answer:

ammonia

Explanation:

Nitrogen fertilizers contain N in the forms of ammonium, nitrate and urea. Upon application to the soil, urea-N rapidly hydrolyzes to ammonia, thus it shares similar characteristics as ammonia-based N fertilizers.

5 0

2 years ago

Fe(no3)3 + na0h > fe(oh)3 + nano3 is balanced or unbalanced?

VashaNatasha [74]

Answer:

Unbalanced

Explanation:

$\text{Fe}(\text{NO}_3)}_3+ \text{NaOH} \rightarrow \text{Fe(OH)}_3 + \text{NaNO}_3\\\\\text{Balanced reaction:}\\\\\text{Fe}(\text{NO}_3)}_3+ 3\text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3\text{NaNO}_3$

5 0

1 year ago

What is the volume of 2 mol of chlorine gas at STP?<br> 2.0 L<br> 11.2 L<br> 22.4 L<br> 44.8 L

nirvana33 [79]

Answer:

44.8 L

Explanation:

Using the ideal gas law equation:

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)

At Standard temperature and pressure (STP);

P = 1 atm

T = 273K

Hence, when n = 2moles, the volume of the gas is:

Using PV = nRT

1 × V = 2 × 0.0821 × 273

V = 44.83

V = 44.8 L

7 0

2 years ago

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔

Vilka [71]

Answer : The correct option is, (C) 1.1

Solution : Given,

Initial moles of $N_2O_4$ = 1.0 mole

Initial volume of solution = 1.0 L

First we have to calculate the concentration $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

The given equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initially c 0

At equilibrium $(c-c\alpha)$ $2c\alpha$

The expression of $K_c$ will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

where,

$\alpha$ = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

$K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}$

$K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}$

$K_c=1.066\aprrox 1.1$

Therefore, the value of equilibrium constant for this reaction is, 1.1

4 0

2 years ago