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svetlana [45]
3 years ago
15

Calculate q (in units of joules) when 1.850 g of water is heated from 22 °C to 33 °C. Report only the numerical portion of your

answer - do not include the symbol for the units.
Chemistry
1 answer:
Minchanka [31]3 years ago
5 0
When q is the heat energy in joules (J)

so, according to this formula, we can get q (in joule unit):

q = M*C*ΔT

when M is the mass of the water sample = 1.85 g

C is the specific heat capacity of water = 4.18 J/g.°C

and Δ T is the difference in temperature (Tf-Ti) = 33 - 22 = 11°C

So, by substitution, we will get the value of q ( in Joule):

∴ q = 1.85 g * 4.18 J/g.°C * 11 °C

      = 85 J
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Could someone help me with this too? Much appreciated! I am a bit stuck.
Mashcka [7]

Answer:

CO

Explanation:

4 0
3 years ago
A sample of helium gas initially at 37.0°C, 785 torr and 2.00 L was heated to 58.0°C while the volume expanded to 3.24 L. What i
spayn [35]

Answer:

0.681 atm

Explanation:

To solve this problem, we make use of the General gas equation.

Given:

P1 = 785 torr

V1 = 2L

T1 = 37= 37 + 273.15 = 310.15K

P2 = ?

V2 = 3.24L

T2 = 58 = 58+273.15 = 331.15K

P1V1/T1 = P2V2/T2

Now, making P2 the subject of the formula,

P2 = P1V1T2/T1V2

P2 = [785 * 2 * 331.15]/[310.15 * 3.24]

P2 = 515.715 Torr

We convert this to atm: 1 torr = 0.00132 atm

515.715 Torr = 515.715 * 0.00132 = 0.681 atm

8 0
3 years ago
Hi May I know how to balance this
almond37 [142]

Answer:

  2Ba₃(PO₄)₂ +6SiO₂ ⇒ P₄O₁₀ +6BaSiO₃

Explanation:

Equating coefficients, you get ...

  aBa₃(PO₄)₂ +bSiO₂ ⇒ cP₄O₁₀ +dBaSiO₃

For Ba: 3a = d

For P: 2a = 4c

For O: 8a +2b = 10c +3d

For Si: b = d

__

Expressing everything in terms of b and c, we get ...

  d = b

  a = b/3 = 2c

From the second, b = 6c, so we have ...

  a = 2c

  b = 6c

  c = c

  d = 6c

And we can write the equation with c=1 as ...

  2Ba₃(PO₄)₂ +6SiO₂ ⇒ P₄O₁₀ +6BaSiO₃

4 0
3 years ago
When 250 ml of water is added to 35 ml of 0.2 M HCl
Paladinen [302]
Assuming that you’re looking for the concentration of water in the solution, then it would be 0.028 M.

You would have to use the formula:
c1v1 = c2v2, where c =concentration and
v = volume

C1 = ?
V1 = 250 mL
C2 = 0.2 M
V2 = 35 mL

C1 x 250 mL = 0.2 M x 35 mL

C1 = (0.2 M x 35 mL) / 250 mL

C1 = 0.028 M of water added to 35mL of 0.2M HCl

Therefore, there is 0.028 M of water added to 35mL of 0.2M HCl
8 0
3 years ago
Read 2 more answers
The equilibrium constant, Kp, for the following reaction is 0.497 at 500K.PCl5(g) PCl3(g) + Cl2(g)If an equilibrium mixture of t
Anestetic [448]

<u>Answer:</u> The equilibrium partial pressure of chlorine gas is 0.360 atm

<u>Explanation:</u>

For the given chemical equation:

PCl_5(g)\rightleftharpoons PCl_3(g)+Cl2(g)

The expression of K_p for above reaction follows:

K_p=\frac{p_{Cl_2}\times p_{PCl_3}}{p_{PCl_5}}

We are given:

K_p=0.497\\p_{PCl_3}=0.651atm\\p_{PCl_5}=0.471atm

Putting values in above equation, we get:

0.497=\frac{p_{Cl_2}\times 0.651}{0.471}\\\\p_{Cl_2}=0.360atm

Hence, the equilibrium partial pressure of chlorine gas is 0.360 atm

5 0
3 years ago
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