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3 years ago

Convert the following to standard notationPLEASE HELP :(

Chemistry

2 answers:

olga55 [171]3 years ago

7 0

1. 1512
2. 0.001510
3. 0.37510
4. 375.10
5. 220000

Hope this helped you

aksik [14]3 years ago

3 0

Answer:

1) 1.5 x 10^3 = 1500 Explanation: move your decimal 3 times to the right ->

2) 1.5 x 10^-3 = 0.0015 Explanation: move your decimal 3 times to the left

3) 3.75 x 10^-2 = 0.0375 Explanation: move your decimal 2 times to the left

4) 3.75 x 10^2 = 375 Explanation: move your decimal to the right twice

5) 2.2 x 10^5 = 220000 Explanation: move your decimal to the right 5 times

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(4.1357 x 10^-15) x (5.4 x 10^2) Answer must be in scientific notation

nadya68 [22]

Answer:

22×10⁻¹³

Explanation:

Scientific notation:

Scientific notation is the way to express the large value in short form.

The number in scientific notation have two parts.

The digits (decimal point will place after first digit)

× 10 ( the power which put the decimal point where it should be)

for example the number 6324.4 in scientific notation will be written as = 6.3244 × 10³

When we multiply or divide the values the number of significant figures must be equal to the less number of significant figures in given value. Thus, in given value,

(4.1357× 10⁻¹⁵) × (5.4×10²)

22×10⁻¹³

5 0

4 years ago

Calculate the concentration of acetic acid and acetate ion in a 0.2 M acetate buffer at pH 5. The p K a of acetic acid is 4.76

Levart [38]

Answer:

[Acetic acid] = 0.07 M

[Acetate] = 0.13 M

Explanation:

pH of buffer = 5

pKa of acetic acid = 4.76

$pH=p_{Ka} + log\frac{[Salt]}{[Acid]}$

Now using Henderson-Hasselbalch equation

$5=4.76 + log\frac{[Acetate]}{[Acetic\;acid]}$

$log\frac{[Acetate]}{[Acetic\;acid]} = 0.24$

$\frac{[Acetate]}{[Acetic\;acid]} = 1.74$ ....... (1)

It is given that,

[Acetate] + [Acetic acid] = 0.2 M ....... (2)

Now solving both the above equations

[Acetate] = 1.74[Acetic acid]

Substitute the concentration of acetate ion in equation (2)

1.74[Acetic acid] + [Acetic acid] = 0.2 M

[Acetic acid] = 0.2/2.74 = 0.07 M

[Acetate] = 0.2 - 0.07 = 0.13 M

5 0

3 years ago

Brandon is traveling in France. He weighs himself at a health club because he is concerned that he has eaten too many croissants

anastassius [24]

Answer:

For a: Yes, he has gained weight.

For b: The number of onions required is 1.

Explanation:

For a:

We are given:

Weight of Brandon checked earlier = 183 lb

Converting this into kilograms, we use the conversion factor:

1 kg = 2.205 lb

So, $183lb\times \frac{1kg}{2.205lb}=83kg$

And, weight of Brandon measured at the health club = 85.0 kg

As, the weight at the health club is more than the previous measured weight. So, Brandon has gained weight.

For b:

We are given:

Amount of diced onions a recipe calls = 125 g

Mass of 1 onion = 4 oz

Converting the mass of 1 onion to ounces, we use the conversion factor:

1 oz = 28.35 g

So, $4oz\times \frac{28.35g}{1oz}=113.4g$

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

$\text{Number of onions}=\frac{\text{Mass of onion a recipe calls}}{\text{Mass of 1 onion}}$

$\text{Number of onions}=\frac{125g}{113.4g}=1.1\approx 1$

Hence, the number of onions required is 1.

6 0

3 years ago

In the reaction below, 22 g of H2S with excess

BartSMP [9]

Answer:

24%

Explanation:

From the question, the limiting reactant is H2S.

The reaction equation is;

2H2S + O2 → 2S + 2H2O

Number of moles of H2S reacted = 22g/34 g/mol = 0.647 moles

According to the reaction equation;

2 moles of H2S yields 2 moles of sulphur

0.647 moles of H2S yields 0.647 moles of sulphur

So;

Theoretical yield of sulphur = 0.647 moles * 32 g/mol = 20.7 g

Actual yield = 5 g

% yield = actual yield/theoretical yield * 100

% yield =5 g/20.7 g * 100

% yield = 24%

7 0

3 years ago

The element lead (Pb) consists of four naturally occurring isotopes with masses of 203.97302, 205.97444, 206.97587, n 207.97663

Zanzabum

Answer:

The atomic mass of lead is: 207.216 u

Explanation:

data Isotopes mass percent

1 203.97302 1.4

2 205.974444 24.1

3 206.97587 22.1

4 207.97663 52.4

atomic mass = (203.97302x 0.014) + (205.974444 x 0.241) +

(206.97587 x 0.221) + (207.97663 x 0.524)

atomic mass = 2.856 +49.639 + 45.742 + 108.979

atomic mass = 207.216 u

8 0

3 years ago

Convert the following to standard notationPLEASE HELP :(​

Convert the following to standard notationPLEASE HELP :(