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4 years ago

What is the introduction of separating salt and sugar?

1 answer:

4 0

So, you could put filters with holes based on the size of the average gram. Which means sugar grams or salt can fit through.

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When the metallic element sodium combines with the nonmetallic element bromine (Br2), is the product a solid, liquid, or gas at

Luden [163]

Answer: it is a solid at room temperature.

Explanation:

The ions of the two reactants are Na{+} and Br{-}, so the formula of the product is NaBr.

and when the NaCl is a solid a room temperature. and when NaBr has similar properties to NaCl, so NaBr is a solid at room temperature too.

so our chemical formula for the product is NaBr and It's a solid at room temperature.

4 0

3 years ago

A reaction produces 74.10 g Ca(OH)2 after 56.08 g CaO is added to 36.04 g H2O. How should the difference in the masses of reacta

melomori [17]

Cao + H2O ---->Ca(OH)2
Calculate the number of each reactant and the moles of the product
that is
moles = mass/molar mass
The moles of CaO= 56.08g/ 56.08g/mol(molar mass of Cao)= 1mole
the moles of water= 36.04 g/18 g/mol= 2.002moles
The moles of Ca (OH)2=74.10g/74.093g/mol= 1mole

The mass of differences of reactant and product can be therefore
explained as
1 mole of Cao reacted completely with 1 mole H2O to produce 1 mole of Ca(OH)2. The mass of water was in excess while that of CaO was limited

3 0

3 years ago

How many moles of O2 are needed to burn 2.56 moles of CH3OH?

lukranit [14]

Answer:

$n_{O_2}=3.84molO_2$

Explanation:

Hello!

In this case, since the combustion reaction of methanol is:

$CH_3OH+\frac{3}{2} O_2\rightarrow CO_2+2H_2O$

In such a way, since there is 1:3/2 mole ratio between methanol and oxygen, we can compute the moles of oxygen that are needed to burn 2.56 moles of methanol as shown below:

$n_{O_2}=2.56molCH_3OH*\frac{\frac{3}{2}molO_2}{1molCH_3OH} \\\\n_{O_2}=3.84molO_2$

Best regards!

6 0

3 years ago

15. What volume of CCI, (d = 1.6 g/cc) contain

anastassius [24]

Answer:

$\boxed{\text{(3) 9.6 L}}$

Explanation:

1. Moles of CCl₄

$n = 6.02 \times 10^{25} \text{ molecules} \times \dfrac{\text{1 mol}}{6.022 \times 10^{23}\text{ molecules}} = \text{100.0 mol}$

2. Molar mass of CCl₄

MM = 1 × 12.01 + 4 × 35.5 = 12.01 + 142 = 154.0 g/mol

3. Mass of CCl₄

$m =\text{100.0 mol} \times \dfrac{\text{154.0 g}}{\text{1 mol}} = \text{15 400 g}$

4. Volume of CCl₄

$V = \text{15 400 g} \times \dfrac{\text{1 cm}^{3}}{\text{1.6 g}} = \text{9600 cm}^{3}\\\\V = \text{9600 cm}^{3} \times \dfrac{\text{1 L}}{\text{1000 cm}^{3}} = \mathbf{{9.6 L}}\\\\\text{The volume of CCl$_{4}$ is } \boxed{\textbf{9.6 L}}$

4 0

3 years ago

A reaction involves the parts of two compounds switching places to form two new compounds. What type of reaction is this?

In-s [12.5K]

Double Displacement would be correct

5 0

4 years ago