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zalisa [80]
3 years ago
14

A fish company delivers 22 kg of salmon, 5.5 kg of crab and 3.48 kg of oysters to your seafood restaurant. What is the mass, in

kilograms of the seafood? What is the total number of pounds?
Chemistry
1 answer:
Rainbow [258]3 years ago
8 0

Answer:

Mass of sea food = 30.98 Kg

Mass of sea food in pound = 68.31 lbs

Explanation:

Salmon, crab and oysters all are sea food.

Mass of sea food = Mass of salmon + Mass of crab + mass of oyster

Mass of salmon = 22 kg

Mass of crab = 5.5 kg

Mass of oysters = 3.48 kg

Mass of sea food = Mass of salmon + Mass of crab + mass of oyster

                             = 22 + 5.5 + 3.48

                             = 30.98 Kg

1 Kg = 2.205 lbs

Therefore, 30.98 kg = 30.98 × 2.205

                                 = 68.31 lbs

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Among the alkali earth metals, the tendency to react with other substances
worty [1.4K]
The answer would be letter A.Among the alkali earth metals, the tendency to react with other substances <span>increases from bottom to top within the group. We know that when there is an increase of metallic property, there will also be an increase of reactivity. </span>
8 0
4 years ago
(15 points). The oxidation of glucose provides the principal energy source for animal cells. The reactants are glucose [C6H12O6(
miss Akunina [59]

Answer:

Check the explanation

Explanation:

The balanced reaction

C6H12O6(s) + 6O2(g) = 6CO2(g) + 6H2O(l)

Standard heat of reaction

Hrxn = 6*Hf(CO2) + 6*Hf(H2O) - 6*Hf(O2) - Hf(C6H12O6)

= 6*(-393.5) + 6*(-285.8) - 6*(0) - (-1274.4)

= - 2801.4 kJ/mol

Part b

Energy consumed by a person = 150 kJ/kg x 57 kg = 8550 kJ

Moles of glucose required = 8550 kJ / (2801.4 kJ/mol)

= 3.052 mol

Mass of glucose required = moles x molecular weight

= 3.052 mol x 180.156 g/mol

= 549.84 g

Part c

1 person requires = 3.052 mol

275 million person require = 275*10^6*3.052 = 8.39 x 10^8 mol

From the stoichiometry of the reaction

1 mol glucose produces = 6 mol CO2

8.39 x 10^8 mol glucose produces = 6*8.39*10^8

= 5.036 x 10^9 mol CO2

Mass of CO2 produced = moles x molecular weight

= 5.036 x 10^9 mol x 44 g/mol

= 2.22 x 10^11 g x 1kg/1000g

= 2.22 x 10^8 kg x 1million/10^6

= 222 million kg

5 0
4 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!

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3 years ago
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When it is acted upon by an outside force
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Read 2 more answers
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Methane is CH4
Ethane is. C2H6

Method by ☆cracking alkanes☆

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Cracking alkanes is the method used to split a long alkane structure into smaller alkanes using high temperature and a catalyst.

e.g. C2H6 -> CH4 + CH2


8 0
3 years ago
Read 2 more answers
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