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3 years ago

If D+2 would react with E-1, what do you predict to be the formula?

Chemistry

1 answer:

harkovskaia [24]3 years ago

8 0

Answer:

DE2

Explanation: for every one D+2 you need two E-1 because +2=-2

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A 100 gram glass container contains 200 grams of water and 50.0 grams of ice all at 0°c. a 200 gram piece of lead at 100°c is ad

ASHA 777 [7]

$0 \; \textdegree{\text{C}}$

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

$E(\text{fusion}) = m(\text{ice}) \cdot L_{f}(\text{water}) = 66.74 \; \text{kJ}$ and
$m(\text{water, final}) = m(\text{water, initial}) + m(\text{ice, initial}) = 0.250 \; \text{kg}$

Let the final temperature of the system be $t \; \textdegree{\text{C}}$ . Thus $\Delta T (\text{water}) = \Delta T (\text{beaker}) = t(\text{initial}) - t_{0} = t \; \textdegree{\text{C}}$

$Q(\text{water}) &= &c(\text{water}) \cdot m(\text{water, final}) \cdot \Delta T (\text{water})= 1.047 \cdot t\; \text{kJ}$ (converted to kilojoules)
$Q(\text{container}) &= &c(\text{glass}) \cdot m(\text{container}) \cdot \Delta T (\text{container})= 0.0837 \cdot t \; \text{kJ}$
$Q(\text{lead}) &= &c(\text{lead}) \cdot m(\text{lead}) \cdot \Delta T (\text{lead})= 0.0255 \cdot (100 - t)\; \text{kJ}$

The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable $t$ .

$E(\text{absorbed} ) = E(\text{released})$

$E(\text{absorbed} ) = E(\text{fushion}) + Q(\text{water}) + Q(\text{container})$
$E(\text{released}) = Q(\text{lead})$

Confirm the uniformity of units, equate the two expressions and solve for $t$ :

$66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)$

$t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}}$ which goes against the initial assumption. Implying that the final temperature does not go above the melting point of water- i.e., $t \le 0 \; \textdegree{\text{C}}$ . However, there's no way for the temperature of the system to go below $0 \; \textdegree{\text{C}}$ ; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at $0 \; \textdegree{\text{C}}$ ; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

3 0

3 years ago

The equation for density, d, is

ss7ja [257]

Answer: The density of substance is $3.38g/cm^3$

Explanation:

To calculate density of a substance, we use the equation:

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

We are given:

Mass of substance = 61.6 g

Volume of substance = $18.2cm^3$

Putting values in above equation, we get:

$\text{Density of substance}=\frac{61.6g}{18.2cm^3}\\\\\text{Density of substance}=3.38g/cm^3$

Hence, the density of substance is $3.38g/cm^3$

8 0

2 years ago

Which cannot be chemically broken down into simpler substances?

Anna71 [15]

Answer:

Explanation:

elements

6 0

2 years ago

Read 2 more answers

Un ácido nítrico tiene una densidad de 10% ,calcular el peso de 1000 centímetros cúbicos

devlian [24]

D = M/V
D = 0.10
V = 1000 cm^3

0.10 = M / 1000
M = 100

Porque la densidad es 10%, se puede usar 0.10 en este formulario para calcular el peso. No sé lo que es la unidad para el peso pero es 100.

4 0

2 years ago

The heat of fusion for water is 80. cal/g. How many calories of heat are needed to melt a 35 g ice cube that has a temperature o

HACTEHA [7]

Answer: 2800 calories

Explanation:

Latent heat of fusion is the amount of heat required to convert 1 mole of solid to liquid at atmospheric pressure.

Amount of heat required to fuse 1 gram of water = 80 cal

Mass of ice given = 35 gram

Heat required to fuse 1 g of ice at $0^0C$ = 80 cal

Thus Heat required to fuse 35 g of ice = $\frac{80}{1}\times 35=2800cal$

Thus 2800 calories of energy is required to melt 35 g ice cube

6 0

3 years ago