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2 years ago

A chemical equation is balanced when the number of each

Chemistry

1 answer:

Lelu [443]2 years ago

4 0

Answer: Atom/element

Explanation: Since matter cannot be created or destroyed, the number of atoms has to be equal on both sides of the equation.

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If I have a 78c and I fail a test that’s worth 15% of my grade what would it drop to ?

koban [17]

Depends on the grade of the test. But for the most part if it’s below a 78, it’ll drop, If it’s above it’ll raise. It shouldn’t drop or raise it more than 3 point/percent.

8 0

2 years ago

A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then

Ede4ka [16]

<u>Answer:</u> The mass of glucose in final solution is 1.085 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ ......(1)

Given mass of glucose = 15.5 g

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Molarity of glucose solution}=\frac{15.5\times 1000}{180.2\times 100}\\\\\text{Molarity of glucose solution}=0.860M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.860M\\V_1=35.0mL\\M_2=?M\\V_2=0.500L=500mL$

Putting values in above equation, we get:

$0.860\times 35.0=M_2\times 500\\\\M_2=\frac{0.860\times 35.0}{500}=0.0602M$

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of glucose solution = 0.0602 M

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0602=\frac{\text{Mass of glucose solution}\times 1000}{180.2\times 100}\\\\\text{Mass of glucose solution}=\frac{0.0602\times 180.2\times 100}{1000}=1.085g$

Hence, the mass of glucose in final solution is 1.085 grams

4 0

3 years ago

Given that w for water is 2. 4×10−14 m^2 at 37 °C. Calculate the ph of a neutral aqueous solution at 37 °c, which is the normal

alexgriva [62]

The pH of a neutral aqueous solution at 37°C is 6.8.

Kw is defined as the dissociation, which is also known as self-ionization, constant of water. this is an equilibrium constant, and its expression is:

Kw = [OH⁻] . [H₃O⁺]

Neutral pH determines that the concentrations of OH⁻ and H₃O⁺ are equal.

<h3>Calculation</h3>

Let us suppose concentration of OH and H₃O⁺ is x, to calculate it:

Kw =[OH⁻] . [H₃O⁺] = x²

x² = 2.4 × 10⁻¹⁴ M²

x = 1.5919 × 10⁻⁷ M

Hence, the concentration of OH and H₃O⁺ (x) = [H₃O⁺] = [OH⁻] = 1.5919×10⁻⁷ M

pH = -log[H₃O⁺] = -log( 1.5919×10⁻⁷ M)

pH = 6.8

Thus, we find that the pH of a neutral aqueous solution at 37 °c (which is the normal human body temperature) is 6.8.

learn more about pH:

brainly.com/question/9529394

#SPJ4

3 0

1 year ago

What mass, in grams, of sodium sulfate is needed to make 230.5 g of a 3.5 % (m/m) aqueous solution of sodium sulfate?

andreev551 [17]

The percent concentration of a solution can be calculated from; mass of solute /mass of solution * 100. The mass of the solute here is 8.1 g.

<h3>What is concentration?</h3>

The term concentration refers to the amount of solute presnt in a solution. There are many ways of expressing concentration such as molarity, molality and percentage.

Here;

mass of solution = 230.5 g

Percent of solute = 3.5 %

3.5 = x/ 230.5 * 100

3.5 = 100x/230.5

230.5(3.5) = 100x

x = 230.5(3.5) /100

x = 8.1 g

Learn more about percent concentration: brainly.com/question/202460?

7 0

1 year ago

While exploring on your grandfathers farm, you find a layer of charcoal that might represent a campfire built by Native American

Ierofanga [76]

Carbon Dating would tell you the age of the charcoal

8 0

3 years ago