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3 years ago

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What is the strongest base that is present after methyl magnesium bromide (CH3MgBr) is treated with water?

1 answer:

sergeinik [125]3 years ago

7 0

Answer:

Explanation:Explanation:the nucleophile attack hydrogen on water causing the single to be unstable as a result of CH4 formation plus OH

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Mr. McPhillip has a ribbon of magnesium (Mg), a silvery metal. As a demonstration for his class, he sets the metal on fire. It b

OLEGan [10]

A. Magnesium is an element and after it is burned it is a compound

5 0

3 years ago

Read 2 more answers

In the following reaction, how many grams of silver can be produced from 49.1 g copper?

notsponge [240]

166.4 g Ag grams of silver can be produced from 49.1 g of copper.

<h3>What is a mole?</h3>

A mole is a very important unit of measurement that chemists use. A mole of something means you have 602,214,076,000,000,000,000,000 of that thing, like how having a dozen eggs means you have twelve eggs.

$1 mol Cu + 2 mol AgNO_3$ → $2 mol Ag + 1 mol Cu(NO3)2$

63.55 g Cu —> 2 x 107.688 g Ag

63.55 g Cu gives 215.376 g of Ag

So, 49.1 g Cu —> $\frac{215.376 g X 49.1 g}{63.55 g}$

= 166.4 g Ag

Hence, 166.4 g Ag grams of silver can be produced from 49.1 g of copper.

Learn more about moles here:

brainly.com/question/26416088

#SPJ1

5 0

2 years ago

What are some examples of gases at room temperature

Lady_Fox [76]

Answer:

Nitrogen, Hydrogen, Oxygen, Chlorine, and Fluorine are all gases at room temperature.

Explanation:

6 0

3 years ago

How much heat is lost when 575 grams molten iron at 1825 k becomes solid iron at 293 k? The melting point or iron is 1811 k.

Galina-37 [17]

Answer:

146 kJ

Explanation:

There are two heat flows in this question.

Heat lost on cooling + heat lost on solidifying = 0

q₁ + q₂ = 0

mCΔT + nΔHsol = 0

Data:

m = 575 g

C = 0.449 J·K⁻¹g⁻¹

T_i = 1825 K

T_f = 1811 K

ΔHsol = -13.8 kJ·mol⁻¹

Calculations:

(a) Heat lost on cooling

ΔT = T_f - T_i = 1811 K - 1825 K = -14 K

q₁ = mCΔT = 575 g × 0.449 J·K⁻¹g⁻¹ × (-14 K) = -361 J = -3.61 kJ

(b) Heat lost on solidifying

$n = \text{575 g} \times \dfrac{\text{1 mol}}{\text{55.84 g}} = \text{10.30 mol}\\\\q_{2} = n\Delta_{\text{sol}}H = \text{10.30 mol} \times \dfrac{\text{-13.8 kJ}}{\text{1 mol}}= \text{-142.1 kJ}$

(c) Total heat lost

q = q₁ + q₂ = -3.61 kJ - 142.1 kJ = -146 kJ

The heat lost was 146 kJ.

5 0

3 years ago

Consider an insulated container in which water (2.58 g) solidifies at 0 oC and 1 atm on the surface of a 55 g aluminum block. As

yarga [219]

Answer:

17.41°C is the change in temperature of the aluminum block.

Explanation:

Latent heat of fusion water = 334 J/g

Heat required to freeze 1 gram of water = -334 J

Heat required to freeze 2.58 grams of water = Q

$Q=2.58 g\times (-334 J/g)=-861.72 J$

Heat lost by water will be equal to heat gained by the aluminum block

$Q'=-Q$

$Q'= -(-861.72 J)=861.72 J$

Formula used to calculate heat absorbed by substance:

$Q=mc\Delta T$

Where:

Q = heat absorbed by substance

c = specific heat of substance

m = Mass of the substance

ΔT = change in temperature of the substance

We have :

Mass of aluminum block= m= 55 g

Heat capacity of aluminum = c = 0.9 J/g°C

Change in temperature of the aluminum block= ΔT = ?

Heat absorbed by the aluminum = Q'

$861.72 J=55 g\times 0.9 J/g^oC\times \Delta T$

$\Delta T=17.41 ^oC$

17.41°C is the change in temperature of the aluminum block.

8 0

4 years ago