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Vanyuwa [196]
3 years ago
6

Give me formula for

Chemistry
2 answers:
jasenka [17]3 years ago
8 0
AlCl₃

CuSO₄

Zn(NO₃)₂

(NH₄)₂CO₃
sleet_krkn [62]3 years ago
7 0

Answer:

1. AlCl₃

2. CuSO4

3. Zn(NO₃)₂

4. (NH4)2CO3

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The amount of gravitational force impacts the weight of objects.<br> true or false
ElenaW [278]

Answer:

True

Explanation:

It's how gravity works.

8 0
3 years ago
What is the ratio of hydrogen(H) to oxygen(O) in hydrogen peroxide(H2O2)?
Alona [7]

Answer: 2:2 but if simplified it’s 1:1

Explanation:

7 0
3 years ago
Need help fast please 15 points!!!
In-s [12.5K]

Answer:

C. disposition and condensation ​

Explanation:

8 0
3 years ago
The dissolution of 0.200 l of sulfur dioxide at 19 °c and 745 mmhg in water yields 500.0 ml of aqueous sulfurous acid. The solut
aivan3 [116]

Answer:

Molarity=1.22\ M

Explanation:

Given:  

Pressure = 745 mm Hg

Also, P (mm Hg) = P (atm) / 760

Pressure = 745 / 760 = 0.9803 atm

Temperature = 19 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (19 + 273.15) K = 292.15 K  

Volume = 0.200 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9803 atm × 0.200 L = n × 0.0821 L.atm/K.mol × 292.15 K  

⇒n = 0.008174 moles

From the reaction shown below:-

H_2SO_3+2NaOH\rightarrow Na_2SO_3+2H_2O

1 mole of H_2SO_4 react with 2 moles of NaOH

0.008174 mole of H_2SO_4 react with 2*0.008174 moles of NaOH

Moles of NaOH = 0.016348 moles

Volume = 13.4 mL = 0.0134 L ( 1 mL = 0.001 L)

So,

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity=\frac{0.016348}{0.0134}\ M

Molarity=1.22\ M

8 0
3 years ago
Predict the boiling point of water at a pressure of 1.5 atm.
Lina20 [59]

Answer:

100.8 °C

Explanation:

The Clausius-clapeyron equation is:

ln\frac{P_{1} }{P_{2}} =-Δ\frac{H_{vap}}{r} (\frac{1}{T_{2}}-\frac{1}{T_{1}}  )

Where 'ΔHvap' is the enthalpy of vaporization; 'R' is the molar gas constant (8.314 j/mol); 'T1' is the temperature at the pressure 'P1' and 'T2' is the temperature at the pressure 'P2'

Isolating for T2 gives:

T_{2}=(\frac{1}{T_{1}} -\frac{Rln\frac{P_{2}}{P_{1}} }{Delta H_{vap}}

(sorry for 'deltaHvap' I can not input symbols into equations)

thus T2=100.8 °C

7 0
3 years ago
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