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attashe74 [19]
3 years ago
6

Need help with chemistry.

Chemistry
2 answers:
Paraphin [41]3 years ago
6 0

Answer:

the second one

Explanation:

hope this helps

Sveta_85 [38]3 years ago
3 0

Answer:

2, Argon PLEASE VOTE 5.0 MARK ME BRAINLIEST AND THANK ME

Explanation:

2,Argon because its only logic answer PLEASE VOTE 5.0 MARK ME BRAINLIEST AND THANK ME

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Who is the first person to discover the periodic table​
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Read 2 more answers
A food is initially at a moisture content of 90% dry basis. Calculate the moisture content in wet basis
Anvisha [2.4K]

Answer:

Moisture content in wet basis = 47.4 %    

Explanation:

<u>Moisture content expresses the amount of water present in a moist sample.</u> <u>Dry basis and wet basis are widely used to express moisture content.   </u>

The next equation express the moisture content in wet basis:  

                MC_{wb}=\frac{MC_{db}}{1+MC_{db}}

<em>where,  MC_{wb}: moisture content in wet basis and  </em>

<em>              MC_{db}: moisture content in dry basis    </em>

We now calculate the moisture content in wet basis:

MC_{wb}=\frac{MC_{db}}{1+MC_{db}}

MC_{wb}=\frac{0.90}{1+0.90}

MC_{wb}= 0.474 = 47.4 % wet basis    

Have a nice day!

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4 years ago
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Rechargeable batteries help save the earth from bad toxins that are thrown into landfills
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3 years ago
Please answer, this is due in 30 minutes
notsponge [240]

Answer:

0.591 g of magnesium phosphate is the theoretical yield.

Magnesium nitrate is the limiting reactant.

Explanation:

Hello!

In this case, since the balanced reaction turns out:

3Mg(NO_3)_2+2Na_3PO_4\rightarrow Mg_3(PO_4)_2+6NaNO_3

Next, we compute the grams of magnesium phosphate yielded by each reactant, considering the present mole ratios and molar masses:

m_{Mg_3(PO_4)_2}^{by\ Mg(NO_3)_2}=1.00gMg(NO_3)_2*\frac{1molMg(NO_3)_2}{148.31gMg(NO_3)_2}*\frac{1molMg_3(PO_4)_2}{3molMg(NO_3)_2}  *\frac{262.86gMg_3(PO_4)_2}{1molMg_3(PO_4)_2} \\\\m_{Mg_3(PO_4)_2}^{by\ Mg(NO_3)_2}= 0.591gMg_3(PO_4)_2\\\\m_{Mg_3(PO_4)_2}^{by\ Na_3PO_4}=1.00gNa_3PO_4*\frac{1molNa_3PO_4}{163.94gNa_3PO_4}*\frac{1molMg_3(PO_4)_2}{2molNa_3PO_4}  *\frac{262.86gMg_3(PO_4)_2}{1molMg_3(PO_4)_2} \\\\m_{Mg_3(PO_4)_2}^{by\ Na_3PO_4} = 0.802gMg_3(PO_4)_2

Thus, we infer that the correct theoretical yielded mass is 0.591 g as magnesium nitrate is the limiting reactant for which it produces the fewest grams of product.

However, is not possible to compute the percent yield since no actual yield is given, and must be provided or indicated by the problem or an experiment and it not here, nevertheless, you may compute the percent yield by dividing the actual yield by the theoretical and then multiplying by 100:

Y=\frac{actual}{0.591g}*100\%

Best regards!

4 0
3 years ago
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