Consider the following incomplete reaction. Mg + 2Y ---> MgCl₂ + H₂ Choose the formula for the missing substance Y.

Why are familiar objects such as pens and paper clips not commonly counted in moles?

Anton [14]

Well ask yourself why don't we count it in moles and you should get your answer.

6 0

2 years ago

If 1.0 mole of CH4 and 2.0 moles of Cl2 are used in the reaction CH4 + 4Cl2 => CCl4 + 4HCl then which of these statements is

nikklg [1K]

Answer:

B,C,D

Explanation:

The yield of CCl4 depends on the amount of CH4 in a 1:1 ratio. The amount of Cl2 is twice that of CH4 hence some must be left over. To ensure that all the Cl2 is used up, more CH4 must added to the system.

4 0

2 years ago

Describe an experience you've had making or building something where the amount of each ingredient or building block came in fix

nignag [31]

Baking a cake is an example of making something where the ingredients must be in fixed ratios. Recipes call for specific ratios of ingredients in order to cook properly, and when a recipe for a cake is modified to feed greater or fewer people the ratio remains the same as the original recipe.

4 0

2 years ago

4.82 g of an unknown metal is heated to 115.0∘C and then placed in 35 mL of water at 28.7∘C, which then heats up to 34.5∘C. What

nikitadnepr [17]

Answer: The specific heat of metal is 2.34 J/g°C

Explanation:

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

Volume of water = 35 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{35mL}\\\\\text{Mass of water}=(1g/mL\times 35mL)=35g$

When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of metal = 4.82 g

$m_2$ = mass of water = 35 g

$T_{final}$ = final temperature = 34.5°C

$T_1$ = initial temperature of metal = 115°C

$T_2$ = initial temperature of water = 28.7°C

$c_1$ = specific heat of metal = ?

$c_2$ = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

$4.82\times c_1\times (34.5-110)=-[35\times 4.186\times (34.5-28.7)]$

$c_1=2.34J/g^oC$

Hence, the specific heat of metal is 2.34 J/g°C

4 0

2 years ago

mastering chem If the experiment below is run for 60 s, 0.16 mol A remain. Which of the following statements is or are true? At

Dvinal [7]

Answer: All of the statements are true.

Explanation:

(a) Considering the system mentioned in the equation:-

The sum of total moles in the flask will always be equal to 1 which leads to confirmation of this statement as for 60 secs= 0.16 mol A and 0.84 mol B

(b) 0<t< 20s, mole A got reduced from 1 mole to 0.54 moles while at 40s to 60s A got decreased from 0.30 moles to 0.16 moles.

0 to 20s is 0.46 (1 - 0.54 = 0.46)mol whereas,

40 to 60s is 0.14 (0.30-.16 = 0.14) mol

(0.46 > 0.14) mol leading this statement to be true as well.

(c) Average rate from t1 = 40 to t2 = 60 s is given by:

$\delta moles/\delta time = 0.30-0.16/60-40 = 0.007 Mol/s$ which is true as well

7 0

2 years ago