Ask question

Ask question

All categories

3 years ago

11

Light will interact with objects by making them appear opaque, translucent, or transparent. Please select the best answer from t

he choices provided T F

1 answer:

lbvjy [14]3 years ago

4 0

Answer:

True (T)

Explanation:

By absorbing or reflecting or letting light pass through, things become opaque, translucent, or transparent.

You might be interested in

Why is it so difficult to experience the northern lights here in michigan? I need to know noww

Anni [7]

Answer:

Think about where you're located on the globe. That'll tell you your answer

Explanation:

7 0

3 years ago

Calculate the number of moles in 7.41 X 1017 atoms Ag.

jolli1 [7]

Answer:

1.23x10^-6 mole

Explanation:

A clear understanding of Avogadro's hypothesis proved that 1mole of any substance contains 6.02x10^23 atoms. This indicates that 1mole of Ag contains 6.02x10^23 atoms.

Now, 1f 1mole of Ag contains 6.02x10^23 atoms, then Xmol of Ag will contain 7.41x10^17 atoms i.e

Xmol of Ag = 7.41x10^17/6.02x10^23 = 1.23x10^-6 mole

4 0

3 years ago

A student mixes chemicals A and B together and records the amount of time it takes for a color change to occur. The student repe

k0ka [10]

Answer:

It is mentioned that the student is mixing chemicals A and B and observes the time taken for the color to change. However, in the experiment, it is noticed that the student has repeated the procedure five times and each time he or she is modifying the concentration of chemical B. Thus, it is clear that the concentration of chemical B is the independent variable in the experiment. An independent variable is illustrated as the variable, which is controlled or modified in the experiment.

8 0

3 years ago

How many grams of MgO are needed to produce 264.0 grams of Mg(OH)2?

sergij07 [2.7K]

15396 g
tell me if its correct

7 0

3 years ago

A 44.0 g sample of an unknown metal at 99.0 oC was placed in a constant-pressure calorimeter of negligible heat capacity contain

tatiyna

Answer:

$C_m=0.474\frac{J}{g\°C}$

Explanation:

Hello.

In this case, since this is a system in which the water is heated up and the metal is cooled down in a calorimeter which is not affected by the heat lose-gain process, we can infer that the heat lost by the metal is gained be water, it means that we can write:

$Q_m=-Q_w$

Thus, in terms of masses, specific heats and temperatures we can write:

$m_mC_m(T_{eq}-T_m)=-m_wC_w(T_{eq}-T_w)$

Whereas the equilibrium temperature is the given final temperature of 28.4 °C and we can compute the specific heat of the metal as shown below:

$C_m=\frac{-m_wC_w(T_{eq}-T_w)}{m_m(T_{eq}-T_m)}$

Plugging the values in and since the density of water is 1.00 g/mL so the mass is 80.0g, we obtain:

$C_m=\frac{-80.0g*4.184\frac{J}{g\°C} (28.4\°C-24.0\°C)}{44.0g(28.4\°C-99.0\°C)}\\\\C_m=0.474\frac{J}{g\°C}$

Best regards!

6 0

2 years ago