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bazaltina [42]
2 years ago
14

Questions: 3Na2SO4 b. How many total atoms of Na are represented in the formula?

Chemistry
1 answer:
Ad libitum [116K]2 years ago
7 0
The compound Na2SO4 contains two sodium atoms, one sulfur atom and four oxygen atoms in its chemical formula.
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Given w = 0, an endothermic reaction has the following.
tamaranim1 [39]

D) + ΔH and +ΔE

Given this is one of the answer choices

8 0
3 years ago
Calculate the heat absorbed by the water in a calorimeter when 175 grams of lead cools from 125.0°C to 22.0°C. The specific heat
Iteru [2.4K]

Answer:

Q = 233.42 J

Explanation:

Given data:

Mass of lead = 175 g

Initial temperature = 125.0°C

Final temperature = 22.0°C

Specific heat capacity of lead = 0.01295 J/g.°C

Heat absorbed by water = ?

Solution:

Heat  absorbed by water is actually the heat lost by the metal.

Thus, we will calculate the heat lost by metal.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 22.0°C - 125.0°C

ΔT = -103°C

Q = 175 g × 0.01295 J/g.°C×-103°C

Q = -233.42 J

Heat absorbed by the water is 233.42 J.

5 0
3 years ago
SO
Burka [1]

Answer:

\large \boxed{\text{E) 721 K; B) 86.7 g}}

Explanation:

Question 7.

We can use the Combined Gas Laws to solve this question.

a) Data

p₁ = 1.88 atm; p₂ = 2.50 atm

V₁ = 285 mL;  V₂ = 435 mL

T₁ = 355 K;     T₂ = ?

b) Calculation

\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}

Question 8. I

We can use the Ideal Gas Law to solve this question.

pV = nRT

n = m/M

pV = (m/M)RT = mRT/M

a) Data:

p = 4.58 atm

V = 13.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 385 K

M = 46.01 g/mol

(b) Calculation

\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}

7 0
3 years ago
In the world of chemistry, a 'mole' is a number of something... a very large number of something.
kipiarov [429]

Given:

No of atoms present= 8.022 x 10^23 atoms

Now we know that 1 mole= 6.022 x 10^23 atoms


Hence number of moles present in 8.022 x 10^23 atoms is calculated as below.

Number of moles

= 8.022 x 10^23/6.022x 10^23

=1.3 moles.

Hence we have 1.3 moles present.

3 0
3 years ago
How is rubidium used
Virty [35]

Answer:

Rubidium is used in vacuum tubes as a getter, a material that combines with and removes trace gases from vacuum tubes. It is also used in the manufacture of photocells and in special glasses. Since it is easily ionized, it might be used as a propellant in ion engines on spacecraft.

Symbol: Rb (37)

Atomic Weight: 85.4678

Atomic Number: 37

Number of Stable Isotopes: 1 (View all isotope .

6 0
3 years ago
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