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2 years ago

What will happen when we add 5 grams of sugar to 100 grams of water? Plz tell

Chemistry

2 answers:

Anarel [89]2 years ago

8 0

<h3>♫ :::::::::::::::::::::::::::::: // Hello There ! // :::::::::::::::::::::::::::::: ♫</h3>

➷ It will most likely dissolve in the water. This is due to the sugar being a soluble substance.

Any further queries, let me know.

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

Effectus [21]2 years ago

8 0

Answer:

mass of solute = 5g

mass of solution = 5g +100g =105g

mass/mass percentage x 100

5/105 x 100 =25 x 100 = 2500

Explanation:

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Find the formula for the hydrate<br>0.737 g MgSO3 and 0.763 g H2O

babunello [35]

The required formula of hydrate is MgSO₃.6H₂O.

<h3>How do we calculate the formula of hydrate?</h3>

The number of moles of water per mole of anhydrous solid (x) will be computed by dividing the number of moles of water by the number of moles of anhydrous solid (x) to find the hydrate's formula.

Moles will be calculated as:
n = W/M, where

W = given mass
M = molar mass

Moles of MgSO₃ = 0.737g / 104.3g/mol = 0.007mol

Moles of H₂O = 0.763g / 18g/mol = 0.04 mol

Number of H₂O molecule = 0.04/0.007 = 5.7 = 6

So formula of hydrate is MgSO₃.6H₂O.

Hence required formula of hydrate compound is MgSO₃.6H₂O.

To know more about hydrate compound, visit the below link:

brainly.com/question/22411417

#SPJ1

6 0

2 years ago

What is the volume of water in 150ml of the 35% of sucrose with a specific gravity of 1.115?

anzhelika [568]

Answer:

The volume of the water is 108.71 mL

Explanation:

Step 1: Data given

Volume of water =150 mL = 0.150 L

concentration of sucrose solution 35 % w/w this means in 100 grams of water we have 35 grams of sucrose

specific gravity =1.115

Step 2: Calculate the density of the solution

Density = specific gravity * density of water

Density of solution = 1.115 * 1g/ mL

Density of solution = 1.115 g/ mL

Step 3: Calculate mass of the solution

Mass of solution = density ¨volume

Mass of solution = 1.115 g/ mL * 150 mL

Mass of solution = 167.25 grams

Step 4: Calculate mass of sucrose

35 % = 0.35 * 167.25 grams

Mass sucrose = 58.54 grams

Step 5: Calculate mass of water

Mass of water = mass of sample - mass of sucrose

Mass of water = 167.25 - 58.54 = 108.71 grams

Step 6: Calculate volume of water

Volume = mass / density

Volume = 108.71 grams / 1g/ mL

Volume = 108.71 mL = 0.10871 L

The volume of the water is 108.71 mL

5 0

3 years ago

Read 2 more answers

A sample of ascorbic acid (vitamin C) is synthesized in the laboratory. It contains 30.0 g carbon and 40.0 g oxygen. Another sam

mote1985 [20]

Answer:

16.933g approximately 17.0 grams

Explanation:

From the simple promotions and given the same compound ascorbic acid (vitamin C)

In the laboratory synthesised ascorbic acid

Mass of carbon = 30.0g

Mas of Oxygen = 40.0g

That is the mass of Oxygen per unit mass of Carbon

Per gram of Carbon we have

(30.0g Carbon)÷30 combines with (40.0g of Oxygen)÷30

That is 4/3g of Oxygen per gram of Carbon

Hence the mass of Oxygen compound that combines with 12.7g of Carbonin natural occurring ascorbic acid (vitamin C) is = 4/3×12.7 = 16.933g approximately 17.0g

8 0

3 years ago

What do elements in thr first two columns of the periodic table have in common?

maxonik [38]

The initial two columns of the periodic table make the s-square, and the components in this square share practically speaking that they have a tendency to lose electrons to pick up soundness.

7 0

3 years ago

Read 2 more answers

An electron in the hydrogen atom makes a transition from an energy state of principal quantum number ni to the n = 2 state. If t

topjm [15]

Answer:

$\boxed{3}$

Explanation:

The Rydberg equation gives the wavelength λ for the transitions:

$\dfrac{1}{\lambda} = R \left ( \dfrac{1}{n_{i}^{2}} - \dfrac{1}{n_{f}^{2}} \right )$

where

R= the Rydberg constant (1.0974 ×10⁷ m⁻¹) and

$\text{$n_{i}$ and $n_{f}$ are the numbers of the energy levels}$

Data:

$n_{f} = 2$

λ = 657 nm

Calculation:

$\begin{array}{rcl}\dfrac{1}{657 \times 10^{-9}} & = & 1.0974 \times 10^{7}\left ( \dfrac{1}{2^{2}} - \dfrac{1}{n_{f}^{2}} \right )\\\\1.522 \times 10^{6} &= &1.0974\times10^{7}\left(\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \right )\\\\0.1387 & = &\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \\\\-0.1113 & = & -\dfrac{1}{n_{f}^{2}} \\\\n_{f}^{2} & = & \dfrac{1}{0.1113}\\\\n_{f}^{2} & = & 8.98\\n_{f} & = & 2.997 \approx \mathbf{3}\\\end{array}\\\text{The value of $n_{i}$ is }\boxed{\mathbf{3}}$

7 0

3 years ago