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nika2105 [10]
3 years ago
13

Which of these contain a polyatomic ion? Mark all that apply.

Chemistry
1 answer:
timama [110]3 years ago
8 0

Answer:

Options d contain polyatomic ion...

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If the mass is 6.5g, and the volume is 25mL.. what is the density? PLEASE ANSWER ASAP
julia-pushkina [17]

Answer:

p = 260 kilogram/cubic meter

Explanation:

ρ =  \frac{m}{V} \\

  = \frac{6.5 g}{25mL}

=  0.26 gram/milliliter

=  260 kilogram/cubic meter

7 0
3 years ago
While 1 gram of fat provides 9 calories, 1 gram of glucose provides 4 calories. Why is that?
seropon [69]

Answer:

fat always has more calories than glucose does

Explanation:

hope hope this helps!

6 0
3 years ago
Why might people add a base to an acidic lake?
spayn [35]

Answer:

To decrease the algae

Explanation:

Acid lakes have more algae than other lakes

6 0
3 years ago
A rectangular field measures 6.0 m by 8.0 m. What is the area of the field in square centimeters (cm 2 )? Use the formula: Area
Lera25 [3.4K]
The answer is C

6.0m x 8.0m= 48.0m

then converting it to cm by multiply
48 x 100
7 0
3 years ago
A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
mamaluj [8]

Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

Molarity=\frac{moles}{\text{Volume of solution(L)}}

Moles of glucose = \frac{18.5 g}{180 g/mol}=0.1028 mol

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

Molarity of the solution = \frac{0.1028 mol}{0.1 L}=1.028 mol/L

A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = M_1=1.208 mol

Volume of the solution taken = V_1=30.0 mL

Molarity of the solution after dilution = M_2

Volume of the solution after dilution= V_2=0.500L = 500 mL

M_1V_1=M_2V_2

M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}

M_2=0.07248 mol/L

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}

Moles of glucose = 0.07248 mol/L\times 0.1 L=0.007248 mol

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.

4 0
4 years ago
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