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3 years ago

1.Convert 2.37 moles of Al(NO3)3 to grams.. Single choice.

Chemistry

1 answer:

aleksley [76]3 years ago

7 0

Answer:

1. 504.8 g Al(NO3)3

2. 14.3 moles O2

Explanation:

g = moles x molar mass = 2.37 x 212.996 = 504.8 g Al(NO3)3

MM Al(NO3)3 = 212.996 g/mol

moles = mass : molar mass = 456.89: 32 = 14.3 moles O2

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if .709 j of heat is added to water and cause the temperature to go up by .036 degrees C what mass of water is present

liberstina [14]

Answer:

0.00471 grams H₂O

Explanation:

To determine the mass, you need to use the following equation:

Q = mcΔT

In this equation,

-----> Q = energy/heat (J)

-----> m = mass (g)

-----> c = specific heat capacity (J/g°C)

-----> ΔT = temperature change (°C)

The specific heat capacity of water is 4182 J/g°C. You can plug the given values into the equation and simplify to isolate "c".

Q = 0.709 J c = 4182 J/g°C

m = ? g ΔT = 0.036 °C

Q = mcΔT <----- Equation

0.709 J = m(4182 J/g°C)(0.036 °C) <----- Insert values

0.709 J = m(150.552) <----- Multiply 4182 and 0.036

0.00471 = m <----- Divide both sides by 150.552

8 0

2 years ago

A 951g sample of carbon steel with a specific heat capacity of .49j is heated to 673k and quenched in oil. The steel cools to 52

nikdorinn [45]

Answer: all I know it’s not -31.5 for ppl taking the k12 test

Explanation: I took the test

5 0

3 years ago

Read 2 more answers

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +

Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

Explanation:

(a) Chemical equation for the given reaction in pure water is as follows.

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial: 0 0

Change: +x +x

Equilibm: x x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

$K_{sp} = [Cu^{+}][Cl^{-}]$

$1.2 \times 10^{-6} = x \times x$

x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$ .

(b) When NaCl is 0.1 M,

$CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$ , $K_{sp} = 1.2 \times 10^{-6}$

$Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$ , $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

$K' = K_{sp} \times K$

= 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial: 0.1 0

Change: -x +x

Equilibm: 0.1 - x x

Now, the equilibrium expression is as follows.

K' = $\frac{CuCl_{2}}{Cl^{-}}$

0.1044 = $\frac{x}{0.1 - x}$

x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$ .

7 0

3 years ago

What would you use to mesuasee the height of the tree

Zielflug [23.3K]

Answer:

if you mean "measure" you can use many things. from inches to meters. sometimes if they are big they are measured in "stories". but the smaller the unit is. the harder it is and longer it takes to measure it

7 0

3 years ago

state the formula for potassium hydroxide and explain, in terms of charges, how you formed the formula

Ganezh [65]

Potassium oxide is an ionic compound. The potassium has a charge of K+ and oxygen has a charge of O2−. We need 2 potassium ions to balance one oxide ion making the formula K2O.

Potassium hydroxide is an ionic compound. The potassium has a charge of K+ and hydroxide has a charge of OH−. We need 1 potassium ion to balance one hydroxide ion making the formula KOH.

K2O+ H2O→KOH

To balance the equation we place a coefficient of 2 in front of the potassium hydroxide.

K2O+H2O→2KOH

I hope this was helpful.

3 0

3 years ago