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Mazyrski [523]
3 years ago
5

How much heat is required to raise the temperature of 20.0g of ice from-12°C to 0°C?

Chemistry
1 answer:
Vladimir79 [104]3 years ago
6 0

Answer:

https://socratic.org/questions/how-much-heat-is-required-to-convert-5-88-g-of-ice-at-12-0-c-to-water-at-25-0-c-

Explanation:

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A liter of milk has a [H] of about 2.51 x 10-7. (You may prefer to think of the hydronium ion concentration, [H3O+], as 2.51 x 1
riadik2000 [5.3K]

Answer:

12-6

Explanation:

7 0
2 years ago
Give a structural formula to match the name: ethylpropylamine
SIZIF [17.4K]
<span>Ethylpropylamine has a chemical formula of C2H13N. It has a molecular weight of 83.166 g/mol. It's considered highly flammable and dangerous if swallowed. It should not come in contact with skin or in eyes and it should not be inhaled.</span>
3 0
3 years ago
Mercury-197 has a half-life of 3 days. Starting
Ahat [919]

Answer:

2.34 gms left

Explanation:

3 weeks = 21 days   =    7 half lives

300 * (1/2)^7 = 2.34 gms

4 0
1 year ago
The equilibrium constant, Kc, for the reaction H2 (g) + I2 (g) ⇄ 2HI (g) at 425°C is 54.8. A reaction vessel contains 0.0890 M H
Mars2501 [29]

Answer: The reaction is not at equilibrium and will proceed to make more products to reach equilibrium.

Explanation:  

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_{eq}

K is the constant of a certain reaction when it is in equilibrium, while Q is the reaction  quotient of activities of products and reactants at any stage other than equilibrium of a reaction.

For the given chemical reaction:

H_2(g)+I_2(g)\rightarrow 2HI(g)

The expression for Q is written as:

Q=\frac{[HI]^2}{[H_2]^1[I_2]^1}

Q=\frac{[0.0890]^2}{[0.215]^1[0.498]^1}

Q=0.074

Given : K_{eq} = 54.8

Thus as Q, the reaction will shift towards the right i.e. towards the product side.

4 0
3 years ago
If 1.0 volumes of 1.0 M solutions of sodium hydroxide and lead (II) nitrate are mized how many moles of product are produced
nadezda [96]

Answer:

0.5 moles of Pb(OH)₂ are produced.

Explanation:    

The reaction is:  

2NaOH + Pb(NO_{3})_{2} \rightarrow Pb(OH)_{2} + 2 NaNO_{3}

If we have 1.0 liter of 1.0 M of sodium hydroxide and lead (II) nitrate, the number of moles are:  

n_{NaOH} = C*V = 1 M*1 L = 1 mol

n_{Pb(NO_{3})_{2}} = C*V = 1 M*1 L = 1 mol

Now, we need to find the limiting reactant knowing that 2 moles of sodium hydroxide react with 1 mol of lead (II) nitrate:

n_{NaOH} = \frac{2 moles_{NaOH}}{1 mol Pb(NO_{3})_{2}}* 1 mol Pb(NO_{3})_{2} = 2 moles

Since we have 1 mol of sodium hydroxide and we need 2 moles to react with lead (II) nitrate, then the limiting reactant is sodium hydroxide.

We can find the number of moles of lead (II) hydroxide produced as follows:

n_{Pb(OH)_{2}} = \frac{1 mol Pb(OH)_{2}}{2 mol NaOH}*1 mol NaOH = 0.5 mol

Therefore, 0.5 moles of Pb(OH)₂ are produced.

I hope it helps you!  

5 0
3 years ago
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