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3 years ago

When naming ionic compounds the ______________ is named first.

Chemistry

1 answer:

grandymaker [24]3 years ago

7 0

The answer is metal. Metals are always named first in ionic compounds, like KNO3 for example. I hope this helps!

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Is a Mercury Thermometer a pure substance or a mixture?

alisha [4.7K]

The mercury inside the thermometer is a pure substance hope this helps :)

5 0

3 years ago

Wafting the air above a chemical is one way to it directly

miskamm [114]

To smell it directly

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4 years ago

Food dye allergies have increased in recent years. One study calculated that each 47.9 g serving of M&M's candy contains 240

77julia77 [94]

Answer:

Mass percent of food dyes = 0.0616%

Explanation:

Given data:

Mass of candy = 47.9 g

Calories = 240

Mass of fat = 10 g

Mass of carbohydrate = 34 g

Mass of protein = 2 g

Mass of food dyes = 29.5 mg

Mass percent of food dyes = ?

Solution:

First of all we will convert the mg into g.

Mass of food dyes = 29.5 mg × 1g /1000 mg = 0.0295 g

Mass percent of food dyes = mass of food dyes / total mass× 100

Now we will put the values.

Mass percent of food dyes = 0.0295 g / 47.9 g × 100

Mass percent of food dyes = 0.000616 × 100

Mass percent of food dyes = 0.0616%

3 0

4 years ago

\What is the electrical charge of an electron? positive negative neutral

sweet [91]

Positive
Negative is negative
It says Electron, remember how the first part of the word starts so like “elect” think of electrical

7 0

4 years ago

If 50 ml of 0.235 M NaCl solution is diluted to 200.0 ml what is the concentration of the diluted solution

Helen [10]

This is a straightforward dilution calculation that can be done using the equation

$M_1V_1=M_2V_2$

where M₁ and M₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and V₁ and V₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.

Here, we have the initial concentration (M₁) and the initial (V₁) and final (V₂) volumes, and we want to find the final concentration (M₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for M₂:

$M_2=\frac{M_1V_1}{V_2}.$

Substituting in our values, we get

$\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].$

So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.

5 0

3 years ago