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2 years ago

2)

Chemistry

1 answer:

Darya [45]2 years ago

7 0

The answer is B. Continental Drift (Confidently)

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Which series of energy transformations occurs as an electric generator powers a light bulb

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Answer:

eletrical energy

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The question is confusing and all the answers are not there but it is B C and D

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Suppose that 0.48 g of water at 25 ∘ C condenses on the surface of a 55- g block of aluminum that is initially at 25 ∘ C . If th

mamaluj [8]

Answer:

$49^oC$

Explanation:

At $25^oC$ , the heat of vaporization of water is given by:

$\Delta H^o_{vap} = 43988 J/mol\cdot \frac{1 mol}{18.016 g} = 2441.6 J/g$

The water here condenses and gives off heat given by the product between its mass and the heat of vaporization:

$Q_1 = \Delta H^o_{vap} m_w$

The block of aluminum absorbs heat given by the product of its specific heat capacity, mass and the change in temperature:

$Q_2 = c_{Al}m_{Al}(t_f - t_i)$

According to the law of energy conservation, the heat lost is equal to the heat gained:

$Q_1 = Q_2$ or:

$\Delta H^o_{vap} m_w = c_{Al}m_{Al}(t_f - t_i)$

Rearrange for the final temperature:

$\Delta H^o_{vap} m_w = c_{Al}m_{Al}t_f - c_{Al}m_{Al}t_i$

We obtain:

$\Delta H^o_{vap} m_w + c_{Al}m_{Al}t_i = c_{Al}m_{Al}t_f$

Then:

$t_f = \frac{\Delta H^o_{vap} m_w + c_{Al}m_{Al}t_i}{c_{Al}m_{Al}} = \frac{2441.6 J/g\cdot 0.48 g + 0.903 \frac{J}{g^oC}\cdot 55 g\cdot 25^oC}{0.903 \frac{J}{g^oC}\cdot 55 g} = 49^oC$

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2 years ago

In a hospital, oxygen is administered to patients at 3.0 atm in a hyperbaric oxygen chamber. Oxygen gas, measuring 600.0 L, is c

atroni [7]

Answer: ifkkk

Explanation:

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3 years ago

a concentration solution of H2so4 is 59.4% by mass (m/m) and has a density of 1.83 g/mL. How many mL of the solution would be re

Blababa [14]

Answer: 41.5 mL

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

$Molarity=\frac{n}{V_s}$

where,

n = moles of solute

$V_s$ = volume of solution in L

Given : 59.4 g of $H_2SO_4$ in 100 g of solution

moles of $H_2SO_4=\frac{\text {given mass}}{\text {molar mass}}=\frac{59.4g}{98g/mol}=0.61$

Volume of solution = $\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml$

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.61\times 1000}{54.6ml}=11.2M$

To calculate the volume of acid, we use the equation given by neutralisation reaction:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of stock acid which is $H_2SO_4$

$M_2\text{ and }V_2$ are the molarity and volume of dilute acid which is $H_2SO_4$

We are given:

$M_1=11.2M\\V_1=mL\\M_2=0.30M\\V_2=1550mL$

Putting values in above equation, we get:

$11.2\times V_1=0.30\times 1550\\\\V_1=41.5mL$

Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid

4 0

2 years ago