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3 years ago

I have no clue what to do, HELP!!

Chemistry

1 answer:

Mashcka [7]3 years ago

3 0

The molar ratios in the reaction are 1:1::1:2, so one Mole (Na2CrO4, produces two moles, (NaNO3).

Using the equation: moles = mass(g) /Mr
1. Calculate the Mr of Na2CrO4
from the periodic table,
Na x 2 = 23 x 2 = 46
Cr x 1 = 52 x 1 = 52
O x 4 = 16 x 4 = 64
46 + 52 + 64 = 162.
moles (na2CrO4) = 253 g / 162 = 1.56 moles.
moles(NaNO3) = 1.56 x 2 = 3.12 moles.

The answer is 3.12 moles.

Hope this helps!
Correct me if I'm wrong.
(Some of this was due to help from Yahoo!)

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Boron sulfide, B2S3(s), reacts violently with water to form dissolved boric acid (H3BO3) and hydrogen sulfide gas

Gwar [14]

The equation structure for the above mentioned reaction can be written as

$B_{2} S_{3}+6 H_{2} O \rightarrow 2 H_{3} B O_{3}+3 H_{2} S \uparrow$

Explanation:

Considering the above reaction, When Boron sulfide, reacts with water more violently to form boric acid and hydrogen sulfide gas.

$B_{2} S_{3}+H_{2} O \rightarrow H_{3} B O_{3}+H_{2} S \uparrow$

In order to balance the equation, we can do as follows.There are 2 B - atoms on both sides of the equation, but only 2 H - atoms, and one O - atom on LHS, so we have to balance it by putting 6 in front of water and 2 in front of Boric acid and 3 in front of hydrogen sulphide gas, so that we have 2 B - atoms, 3 - S atoms, 12 H - atoms on both sides of the equation, and it is balanced. Balanced equation is given as,

$B_{2} S_{3}+6 H_{2} O \rightarrow 2 H_{3} B O_{3}+3 H_{2} S \uparrow$

Thus a Balanced equation of the above mentioned reaction is written.

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3 years ago

Brain if correct (3)

Anni [7]

Answer:

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Explanation:

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8 0

2 years ago

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Consider the unbalanced chemical equation HBr + B = BBr3 + H2. When

Vitek1552 [10]

Answer:

B and BBr3

Explanation:

1) 3HBr + B = BBr3 + H2 (double all equation because H2)

2) 6HBr + 2B = 2BBr3 + 3H2

4 0

3 years ago

How many moles are there in 132g of CO2

vovangra [49]

You can use grams to moles and moles to grams. In your case just grams to moles. So since you're given grams, you would divide that by the molar mass of CO2 because that's how many grams are in one mole. The mass for Carbon is 12.0104 g/mol and Oxygen it's 15.9994 g/mol so to find the molar mass you would add 12.0104 + (2*15.9994) which gives you a molar mass of 44.0095 g/mol. You divide your given mass (132g) by the molar mass, so there's 2.9993 moles or approximately 3 moles in 132 g of CO2.

8 0

3 years ago

A 27 kg iron block initially at 375 C is quenched in an insulated tank that contains 130kg of water at 26 C. Assume the water th

Bess [88]

Solution :

a). Applying the energy balance,

$\Delta E_{sys}=E_{in}-E_{out}$

$0=\Delta U$

$0=(\Delta U)_{iron} + (\Delta U)_{water}$

$0=[mc(T_f-T_i)_{iron}] + [mc(T_f-T_i)_{water}]$

$0 = 27 \times 0.45 \times (T_f - 375) + 130 \times 4.18 \times (T_f-26)$

$t_f=33.63^\circ C$

b). The entropy change of iron.

$\Delta s_{iron} = mc \ln\left(\frac{T_f}{T_i} \right)$

$= 27 \times 0.45\ \ln\left(\frac{33.63 + 273}{375 + 273} \right)$

= -9.09 kJ-K

Entropy change of water :

$\Delta s_{water} = mc \ \ln\left(\frac{T_f}{T_i} \right)$

$= 130 \times 4.18\ \ln\left(\frac{33.63 + 273}{26 + 273} \right)$

= 10.76 kJ-K

So, the total entropy change during the process is :

$\Delta s_{tot} = \Delta s_{iron} + \Delta s_{water}$

= -9.09 + 10.76

= 1.67 kJ-K

c). Exergy of the combined system at initial state,

$X=(U-U_{0}) - T_0(S-S_0)+P_0(V-V_0)$

$X=mc (T-T_0) - T_0 \ mc \ \ln \left(\frac{T}{T_0} \right)+0$

$X=mc\left((T-T_0)-T_0 \ ln \left(\frac{T}{T_0} \right)\right)$

$X_{iron, i} = 27 \times 0.45\left(((375+273)-(12+273))-(12+273) \ln \frac{375+273}{12+273}\right)$

$X_{iron, i} =63.94 \ kJ$

$X_{water, i} = 130 \times 4.18\left(((26+273)-(12+273))-(12+273) \ln \frac{26+273}{12+273}\right)$

$X_{water, i} =-13.22 \ kJ$

Therefore, energy of the combined system at the initial state is

$X_{initial}=X_{iron,i} +X_{water, i}$

= 63.94 -13.22

= 50.72 kJ

Similarly, Exergy of the combined system at initial state,

$X=(U_f-U_{0}) - T_0(S_f-S_0)+P_0(V_f-V_0)$

$X=mc\left((T_f-T_0)-T_0 \ ln \left(\frac{T_f}{T_0} \right)\right)$

$X_{iron, f} = 27 \times 0.45\left(((33.63+273)-(12+273))-(12+273) \ln \frac{33.63+273}{12+273}\right)$

$X_{iron, f} = 216.39 \ kJ$

$X_{water, f} = 130 \times 4.18\left(((33.63+273)-(12+273))-(12+273) \ln \frac{33.63+273}{12+273}\right)$

$X_{water, f} =-9677.95\ kJ$

Thus, energy or the combined system at the final state is :

$X_{final}=X_{iron,f} +X_{water, f$

= 216.39 - 9677.95

= -9461.56 kJ

d). The wasted work

$X_{in} - X_{out}-X_{destroyed} = \Delta X_{sys}$

$0-X_{destroyed} =$

$X_{destroyed} = X_{initial} - X_{final}$

= 50.72 + 9461.56

= 9512.22 kJ

6 0

3 years ago