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3 years ago

How many significant figures are in 5.76 e 15

Chemistry

2 answers:

Dvinal [7]3 years ago

5 0

Answer: 3 significant figures

Explanation: because in scientific notation only the before and after decimal number are considered but in standard form all numbers including exponents are all significant figures

meriva3 years ago

3 0

Answer:

3 significant figures

Explanation:

0's are not significant.

I hope it helps you!

$GraceRosalia$

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P4O10(s) = -3110 kJ/mol H2O(l) = -286 kJ/mol H3PO4(s) = -1279 kJ/mol Calculate the change in enthalpy for the following process:

mash [69]

Answer:

-290KJ/mol

Explanation:

ΔHrxn = ΔHproduct - ΔHreactant

ΔHrxn= 4ΔHH3PO4 - {6ΔHH2O + ΔHP4O10}

ΔHrxn = 4(-1279) - [6(-286) - 3110]

= -5116 -(-1716-3110)

= -5116-(-4826)

= -5116 + 4826 = -290KJ/mol

6 0

3 years ago

? Answer the question below. Type your response in the space provided. What volume of a 2.5 M stock solution of acetic acid (HC2

Novay_Z [31]

Data:
$M_{concentrated} = 2.5\:mol$
$V_{concentrated} = ?$
$M_{dilute} = 0.50\:mol$
$V_{dilute} = 100\:mL\to0.100\:L$

Formula: Dilution Calculations

 $M_{concentrated} * V_{concentrated} = M_{dilute} * V_{dilute}$

Solving:


$M_{concentrated} * V_{concentrated} = M_{dilute} * V_{dilute}$
$2.5 * V_{concentrated} = 0.50 * 0.100$
$2.5V_{concentrated} = 0.05$
$V_{concentrated} = \frac{0.05}{2.5}$
$\boxed{\boxed{V_{concentrated} = 0.02\:L\:or\:20\:mL}} \end{array}}\qquad\quad\checkmark$

3 0

3 years ago

For many purposes we can treat ammonia NH3 as an ideal gas at temperatures above its boiling point of −33.°C. Suppose the temper

Keith_Richards [23]

Answer:

The new pressure will be 0.225 kPa.

Explanation:

Applying combined gas law:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1\text{ and }V_1$ are initial pressure and volume at initial temperature $T_1$ .

$P_2\text{ and }V_2$ are final pressure and volume at initial temperature $T_2$ .

We are given:

$P_1=0.29 kPa\\V_1=V\\P_2=?\\V_2=V+50\% of V=1.5 V$

$T_1=-25^oC=248.15 K$

$T _2 = 16^oC=289.15 K$

Putting values in above equation, we get:

$\frac{0.29 kPa\times V}{248.15 K}=\frac{P_2\times 1.5V}{289.15 K}$

$P_1=0.225 kPa$

Hence, the new pressure will be 0.225 kPa.

4 0

3 years ago

Which statements are true in regard to the VSEPR theory?

Naddik [55]

The correct option is D. 2) Anti-bonding electrons or lone pairs. These lone pairs, and bonds helps to form the shape which keeps these electrons separate as possible.

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2 years ago

Write 151.5 cm to meters in the correct significant figures

iogann1982 [59]

Answer:

1.515m

hope that helps uh :)

3 0

2 years ago

How many significant figures are in 5.76 e 15​

How many significant figures are in 5.76 e 15