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natka813 [3]
3 years ago
9

Could someone help me with this? No links, will give brainliest.

Chemistry
1 answer:
Ganezh [65]3 years ago
6 0

Answer: I think it represents the SiO

Explanation:

You might be interested in
A party balloon contains 5.50 x 10 ^22 atoms of helium gas. What is the mass, in grams of the
eduard

Answer:

\boxed {\boxed {\sf A. \ 0.365 \ g\ He}}

Explanation:

<u>1. Convert Atoms to Moles</u>

We must use Avogadro's Number: 6.022*10²³. This is the number of particles (atoms, molecules, ions, etc.) in 1 mole of a substance. In this case, the particles are atoms of helium. We can create a ratio.

\frac {6.022*10^{23} \ atoms \ He}{1 \ mol \ He}

Multiply by the given number of helium atoms.

5.50 *10^{22} \ atoms \ He *\frac {6.022*10^{23} \ atoms \ He}{1 \ mol \ He}

Flip the fraction so the atoms of helium cancel.

5.50 *10^{22} \ atoms \ He *\frac {1 \ mol \ He}{6.022*10^{23} \ atoms \ He}

5.50 *10^{22} *\frac {1 \ mol \ He}{6.022*10^{23} }

\frac {5.50 *10^{22} \ mol \ He}{6.022*10^{23} }= 0.09133178346 \ mol \ He

<u>2. Convert Moles to Grams</u>

We must use the molar mass, which is found on the Periodic Table.

  • Helium (He): 4.00 g/mol

Use this as a ratio.

\frac { 4.00 \ g \ He }{ 1 \ mol \ He}

Multiply by the number of moles we calculated. The moles will then cancel.

0.09133178346 \ mol \ He *\frac { 4.00 \ g \ He }{ 1 \ mol \ He}

0.09133178346*\frac { 4.00 \ g \ He }{ 1 }

0.3653271338 \ g\ He

<u>3. Round </u>

The original measurement has 3 significant figures (5, 5, and 0). Our answer must have the same. For the number we calculated, it is thousandth place. The 3 in the ten thousandth place tells us to leave the 5.

0.365 \ g\ He

The mass is <u>0.365 grams of helium</u> so choice A is correct.

7 0
2 years ago
0.0278, how many significant figures
My name is Ann [436]

Answer:

There are three significant figures

Explanation:

When counting sig figs you don't count the zeros unless it is between a number greater than zero. The two zeros aren't between the greater numbers so there are only 3.

3 0
3 years ago
What type of circuit is illustrated?
Ber [7]
I believe it's B: series circuit



good luck
5 0
3 years ago
Read 2 more answers
A student obtains the following data: Mass of empty, dry graduated cylinder: 21.577 g Volume added of NaCl solution: 4.602 mL Ma
klasskru [66]

<u>Answer:</u> The density of NaCl solution is 3.930 g/mL

<u>Explanation:</u>

We are given:

Mass of cylinder, m_1 = 21.577 g

Mass of NaCl and cylinder combined, M = 39.664 g

Mass of NaCl, m_2 = (M-m_1)=(39.664-21.577)g=18.087g

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 NaCl = 18.087 g

Volume of NaCl solution = 4.602 mL

Putting values in above equation, we get:

\text{Density of NaCl}=\frac{18.087g}{4.602mL}\\\\\text{Density of NaCl}=3.930g/mL

Hence, the density of NaCl solution is 3.930 g/mL

7 0
3 years ago
Can someone answer all please.
boyakko [2]

Answer:

poszukaj w necie , proszę a nie prosisz innych o gotowca

8 0
3 years ago
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