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3 years ago

PLEASE HURRY IM TIMED I WILL GVE BRAINLIEST IF CORRECT

1 answer:

6 0

So basically D,E They do absorb water but the roots mainly do that so I don’t know if you should put that.
Please give me brainliest if I’m correct

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Jones Soda contains 33g added sugar (C6H12O6) in a 355 ml bottle, while Sierra Mist contains 62g added sugar in a 591 ml bottle.

ArbitrLikvidat [17]

C6H12O6 molar mass: 180.15768 g

solute: sugar

molarity = moles of solute / liters of solution

Jones Soda:

33 g / 180.15768 g = 0.18 moles C6H12O6

M = 0.18 g / 0.355 L

M = 0.52

Sierra Mist:

62 g / 180.15768 g = 0.34 moles C6H12O6

M = 0.34 g / 0.591 L

M = 0.58

Sienna Mist has a higher molarity and is more concentrated.

7 0

3 years ago

If a light bulb is missing or Broken in a parallel circuit,will the other bulb light

umka2103 [35]

Yes, if it’s a parallel circuit the wires are two different wires so it will light because that bulb isn’t connected to the one that went out

5 0

3 years ago

A student reacted NiS2 (MM = 122.83 g/mol) with O2 (MM=32.00 g/mol) to make SO2 (MM=64.07 g/mol) according to this balanced equa

Ilya [14]

Answer:

Percentage yield = 0.49 × 10² %

Explanation:

Given data:

Actual yield of SO₂ = 4.309 ×10² g

Theoretical yield of SO₂ = 8.78 ×10² g

Percentage yield = ?

Solution:

Chemical equation:

2NiS₂ + 5O₂ → 2NiO + 4SO₂

Percentage yield:

Percentage yield = actual yield / theoretical yield × 100

Percentage yield = 4.309 ×10² g / 8.78 ×10² g × 100

Percentage yield = 0.491 × 100

Percentage yield = 49.1%

In scientific notation:

Percentage yield = 0.49 × 10² %

5 0

3 years ago

A student isolated 7.2 g of 1-bromobutane reacting equimolar amounts of 1-butanol (10 ml) and NaBr (11.1 g) in the presence of s

Alla [95]

Answer: The percent yield of the 1-bromobutane is 48.65 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For NaBr:

Given mass of NaBr = 11.1 g

Molar mass of NaBr = 103 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaBr}=\frac{11.1g}{103g/mol}=0.108mol$

The chemical equation for the reaction of 1-butanol and NaBr is:

$\text{1-butanol + NaBr}\rightarrow \text{1-bromobutane}$

By Stoichiometry of the reaction

1 mole of NaBr produces 1 mole of 1-bromobutane

So, 0.108 moles of NaBr will produce = $\frac{1}{1}\times 0.108=0.108$ moles of 1-bromobutane

Now, calculating the mass of 1-bromobutane from equation 1, we get:

Molar mass of 1-bromobutane = 137 g/mol

Moles of 1-bromobutane = 0.108 moles

Putting values in equation 1, we get:

$0.108mol=\frac{\text{Mass of 1-bromobutane}}{137g/mol}\\\\\text{Mass of 1-bromobutane}=(0.108mol\times 137g/mol)=14.80g$

To calculate the percentage yield of 1-bromobutane, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 1-bromobutane = 7.2 g

Theoretical yield of 1-bromobutane = 14.80 g

Putting values in above equation, we get:

$\%\text{ yield of 1-bromobutane}=\frac{7.2g}{14.80g}\times 100\\\\\% \text{yield of 1-bromobutane}=48.65\%$

Hence, the percent yield of the 1-bromobutane is 48.65 %

5 0

3 years ago

Whick ions profuce similar colors in the flame tests?

arlik [135]

Ba2+ and Cu2+, and Sr2+ and Li+

5 0

3 years ago