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Arisa [49]
3 years ago
14

Which substance has a molar mass of 33.99 g/mol?

Chemistry
1 answer:
alex41 [277]3 years ago
6 0

Answer:

PH₃.

Explanation:

Hello.

In this case, among the options:

PH₃, GaF₃, SF₂ and CO₂

We compute their molar masses by adding the atomic masses of the constituting elements by their subscripts:

M_{PH_3}=m_P+3*m_H=30.97+3*1.01=34.00g/mol\\\\M_{GaF_3}=m_{Ga}+3*m_F=69.72+3*19.00=126.72g/mol\\\\M_{SF_2}=m_{S}+2*m_F=32.07+2*19.00=70.07g/mol\\\\M_{CO_2}=m_{C}+2*m_O=12.01+2*16.00=44.01g/mol

Thus, the answer is PH₃.

Best regards.

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3 years ago
If the p-side has a higher doping concentration, explain how to keep tuning to the same radio channel?
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increase in temperature of the intrinsic semiconductor

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A balance in the intrinsic concentration helps in tuning to the same radio channel.

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Explanation:

<em>Hi</em><em> </em><em>there</em><em>!</em><em>!</em>

<em>you</em><em> </em><em>asked</em><em> </em><em>to</em><em> </em><em>multiply</em><em> </em><em>these</em><em> </em><em>all</em><em> </em><em>right</em><em>,</em>

<em>you</em><em> </em><em>can</em><em> </em><em>simply</em><em> </em><em>multiply</em><em> </em><em>it</em><em> </em><em>,</em>

<em>=</em><em>3</em><em>cm</em><em> </em><em>×</em><em> </em><em>4</em><em> </em><em>cm</em><em> </em><em>×</em><em> </em><em>1</em><em>cm</em>

<em>=</em><em> </em><em>1</em><em>2</em><em>cm</em><em>^</em><em>2</em><em>×</em><em>1</em><em>cm</em><em> </em><em> </em><em> </em><em> </em><em>(</em><em>4</em><em>×</em><em>3</em><em>=</em><em>1</em><em>2</em><em>)</em>

<em>=</em><em> </em><em>1</em><em>2</em><em>cm</em><em>^</em><em>3</em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>(</em><em>1</em><em>2</em><em>×</em><em>1</em><em>=</em><em>1</em><em>2</em><em>)</em>

<em>Therefore</em><em>, </em><em> </em><em>the</em><em>answer is</em><em> </em><em>1</em><em>2</em><em> </em><em>cm</em><em>^</em><em>3</em><em>.</em>

<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em>

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