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4 years ago

What is true for a substance that releases energy

Chemistry

1 answer:

hammer [34]4 years ago

6 0

The sun is a true substance that gives off energy.

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WILL GIVE OUT BRAINLY

Margarita [4]

Answer:

3. An oxygen atom with 8 electrons, 8 protons, and 9 neutrons .

Explanation:

if an atom contains equal numbers of protons and electrons, the atom is described as being neutral.

4 0

3 years ago

On the pH scale higher numbers indicate ?

mote1985 [20]

Answer:

Higher numbers indicate bases- not acidic/less acidic

Explanation:

6 0

3 years ago

Read 2 more answers

I am very confused, our teacher didn’t properly teach us this. How do you do it?

Fudgin [204]

Answer:

You multiply the 3 numbers together to get your volume, in this case it would be 4058.488 cm^3 (cm cubed)

V: 4058.488cm^3 ( round up to 4058.5 for convenience)

M: 27579

D: ?

So we divide mass by the volume to get density, which is

27579 / 4058.5 = ~6.79536 (can round up to 7 or 6.8)

This graph can help a lot, so maybe try and memorize it, hope I helped.

3 0

3 years ago

Read 2 more answers

How should the student label each circle in this diagram?

worty [1.4K]

Answer:

Reaction a: chemical, reaction b: nuclear

Explanation:

8 0

3 years ago

What mass of lithium phosphate would you mass to make 2.5 liter of 1.06 M lithium

alex41 [277]

Answer:

Approximately $3.06 \times 10^{2}\; \rm g$ (approximately $306\; \rm g$ .)

Explanation:

Calculate the quantity $n$ of lithium phosphate in $V = 2.5\; \rm L$ of this $c = 1.06\; \rm M = 1.06\; \rm mol \cdot L^{-1}$ lithium phosphate solution.

$\begin{aligned}n &= c \cdot V\\ &= 2.5\; \rm L \times 1.06\; mol \cdot L^{-1}\\ &= 2.65\; \rm mol\end{aligned}$ .

Empirical formula of lithium phosphate: $\rm Li_3PO_4$ .

Look up the relative atomic mass of $\rm Li$ , $\rm P$ ,and $\rm O$ on a modern periodic table:

$\rm Li$ : $6.94$ .
$\rm P$ : $30.974$ .
$\rm O$ : $15.999$ .

Calculate the formula mass of $\rm Li_3PO_4$ :

$M(\rm Li_3PO_4) = 3 \times 6.94 + 30.974 + 4 \times 15.999 = 115.79\; \rm g \cdot mol^{-1}$ .

Calculate the mass of that $n = 2.65\; \rm mol$ of $\rm Li_3PO_4$ formula units:

$\begin{aligned}m &= n \cdot M \\ &= 2.65\; \rm mol \times 115.79\; \rm g\cdot mol^{-1} \\ &\approx 3.06 \times 10^{2}\; \rm g \end{aligned}$ .

5 0

3 years ago