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wel
3 years ago
11

2.5 piece of lithium is dropped into a 500 g sample of water at 22.1 C.ten temperature of the water increases to 23.5 C. How muc

h heat energy is released in the reaction
Chemistry
1 answer:
Dominik [7]3 years ago
3 0

Answer:

2.6 kJ  

Explanation:

The formula for the amount of heat (q) absorbed by the water is

q = mCΔT

1. Calculate ΔT

ΔT = 23.5 °C - 22.1 °C = 1.4 °C

2. Calculate q

q₂ = mCΔT = 500 g × 4.184 J·°C⁻¹g⁻¹ × 1.4 °C = 2900 J = 2.9 kJ

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Determine the expected diffraction angle for the first-order diffraction from the (111) set of planes for FCC nickel (Ni) when m
faust18 [17]

Answer:

56°

Explanation:

First calculate a:

a=2 R \sqrt{2}=2(0.1246) \sqrt{2}=0.352 \mathrm{nm}

The interplanar spacing can be calculated from:

d_{111}=\frac{a}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{0.352}{\sqrt{3}}=0.203 \mathrm{nm}

The diffraction angle is determined from:

\sin \theta=\frac{n \lambda}{2 d_{111}}=\frac{1(0.1927)}{2(0.2035)}=0.476

Solve for \theta

\theta=\sin ^{-1}(0.476)=28^{\circ}

The diffraction angle is:

2 \theta=2\left(28^{\circ}\right)=56^{\circ}

4 0
3 years ago
A gas at constant temperature has a pressure of 404.6 kPa with a volume of 12 ml. If the volume changes to 43ml, what is the new
blagie [28]

Answer:

The answer is

<h2>112.912 kPa</h2>

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new pressure

P_2 =  \frac{P_1V_1}{V_2}  \\

404.6 kPa = 404600 Pa

From the question we have

P_2 =  \frac{404600 \times 12}{43}  =  \frac{4855200}{43}  \\  = 112911.6279... \\  = 112912

We have the final answer as

<h3>112.912 kPa</h3>

Hope this helps you

4 0
3 years ago
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is ________ M sodium ion and ________ M sulfate i
omeli [17]

Answer:

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion.  (option E)

Explanation:

Step 1: Data given

Volume = 500 mL = 0.500 L

The concentration sodium sulfate = 2.104 M

Step 2: The equation

Na2SO4 → 2Na+ + SO4^2-

For 1 mol Na2SO4 we have 2 moles sodium ion (Na+) and 1 mol sulfate ion (SO4^2-)

Step 3: Calculate the concentration of the ions

[Na+] = 2*2.104 M = 4.208 M

[SO4^2-] = 1*2.104 M = 2.104 M

The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion.  (option E)

8 0
3 years ago
The substance in acid rain that makes it more acidic than natural rainwater is _____.
sasho [114]
I’m pretty sure it’s sulphur dioxide
4 0
3 years ago
Read 2 more answers
HELP. I DONT KNOW WHAT TO DO. Barium sulfate, BaSO4, is made by the following reaction.
Ghella [55]

Answer:

\boxed{\text{66.95 g BaSO$_{4}$}}

Explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:         261.34                         233.39

              Ba(NO₃)₂ + Na₂SO₄ ⟶ BaSO₄ + 2NaNO₃

m/g:         75.00

1. Moles of Ba(NO₃)₂

\text{Moles of Ba(NO$_{3})_{2}$} = \text{75.00 g} \times \dfrac{\text{1 mol}}{\text{261.34 g}} = \text{0.286 98 mol}

2. Moles of BaSO₄

The molar ratio is (1 mol BaSO₄/1 mol Ba(NO₃)₂

\text{Moles of BaSO$_{4}$}= \text{0.286 98 mol Ba(NO$_{3})_{2}$ } \times \dfrac{\text{1 mol BaSO$_{4}$}}{\text{1 mol Ba(NO$_{3})_{2}$}} = \text{0.286 98 mol BaSO$_{4}$}

3. Mass of BaSO₄

\text{Mass of BaSO$_{4}$} = \text{0.286 98 mol BaSO$_{4}$} \times \dfrac{\text{233.39 g BaSO$_{4}$}}{\text{1 mol BaSO$_{4}$}} = \textbf{66.98 g BaSO$_{4}$}\\\\\text{The theoretical yield of barium sulfate is } \boxed{\textbf{66.98 g BaSO$_{4}$}}

8 0
3 years ago
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