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3 years ago

15

Which mineral is used in batteries

1 answer:

ankoles [38]3 years ago

4 0

Batteries come in all shapes and sizes, and depending on the type of battery, the minerals that compose them are mainly lithium, cobalt, graphite, nickel and manganese.

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Mature elephants consume an average of 600 pounds of food per day. If an elephant

MAXImum [283]

Answer:

$\large \boxed{\text{4 600 000 kg or 4600 Mg}}$

Explanation:

1. Calculate the adult life

If the elephant is an adult from its 10th birthday until the day before its 56th birthday, its

Adult life = 55 - 10 + 1 = 46 yr

2. Convert years to days

$\text{Adult life} = \text{46 yr} \times \dfrac{\text{365.25 da}}{\text{1 yr}} = \text{16 800 da}$

3. Convert days to pounds of feed

$\text{Feed} = \text{16 800 da} \times \dfrac{\text{600 lb}}{\text{1 da}} = 1.01 \times 10^{7} \text{ lb}$

4. Convert pounds to kilograms and megagrams

$\text{Feed} =1.01 \times 10^{7} \text{ lb} \times \dfrac{\text{1 kg}}{\text{2.20 lb}} =\textbf{4 600 000 kg} = \textbf{4600 Mg}\\\\\text{The elephant will eat $\large \boxed{\textbf{4 600 000 kg or 4600 Mg}}$ of food.}$

Note: The answer can have only two significant figures because that is all you gave for age of the elephant.

5 0

4 years ago

1. Alexa and her family travelled 5 hours south east on 1-45 to crystal

Gre4nikov [31]

Answer:

Average speed = 68 mph

Explanation:

Given that,

Total distance traveled by the family, d = 340 miles

The family traveled 5 hours southeast.

We need to find her average speed. The speed of an object is given by the total distance covered divided by time taken. So,

$v=\dfrac{d}{t}$

Put all the values,

$v=\dfrac{340}{5}\\\\v=68\ mph$

So, her average speed is equal to 68 mph.

5 0

3 years ago

How many grams are in 4 moles of aluminum phosphate

vladimir2022 [97]

Answer:

487.8116

Explanation:

4 0

4 years ago

Read 2 more answers

In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691

andre [41]

Explanation:

Let us assume that the value of $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s$

Also at 1500 K, $K_{r}$ = $2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s$

$K_{r} = 400.613 m^{6}/mol^{2}s$

Relation between $K_{p}$ and $K_{c}$ is as follows.

$K_{p} = K_{c}RT$

Putting the given values into the above formula as follows.

$K_{p} = K_{c}RT$

$0.003691 = K_{c} \times 8.314 \times 1500$

$K_{c} = 2.9 \times 10^{-7}$

Also, $K_{c} = \frac{K_{f}}{K_{r}}$

or, $K_{f} = K_{c} \times K_{r}$

= $2.9 \times 10^{-7} \times 400.613$

= $1.1617 \times 10^{-4} m^{6}/mol^{2}s$

Thus, we can conclude that the value of $K_{f}$ is $1.1617 \times 10^{-4} m^{6}/mol^{2}s$ .

6 0

3 years ago

An element has a half-life of 30 years. if 1.0 mg of this element decays over a period of 90 years, how many mg of this element

labwork [276]

Answer:

144.6

Explanation:

6 0

3 years ago