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3 years ago

WILL GIVE BRAINLIEST IF SOMEONE ANSWERS THIS CORRECTLY ASAP!!!!

Chemistry

1 answer:

dybincka [34]3 years ago

3 0

Answer:Chemical change

Explanation: because a substance in a state different from the state of the original substances is formed

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AlekseyPX

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2 years ago

The stomach and intestines are organs of the body system which _______.

yulyashka [42]

Answer:

A. Digests food

Explanation:

Look about diggestive process in Google

Best regards

3 0

3 years ago

Read 2 more answers

You need to make 10 mL of 2 mg/ml solution of protein and you have 25 mg/mL solution. How much protein solution and water do you

meriva

Answer:

0.8 mL of protein solution, 9.2 mL of water

Explanation:

The dilution equation can be used to relate the concentration C₁ and volume V₁ of the stock/undiluted solution to the concentration C₂ and volume V₂ of the diluted solution:

C₁V₁ = C₂V₂

We would like to calculate the value for V₁, the volume of the inital solution that we need to dilute to make the required solution.

V₁ = (C₂V₂) / C₁ = (2mg/mL x 10mL) / (25 mg/mL) = 0.8 mL

Thus, a volume of 0.8 mL of protein solution should be diluted with enough water to bring the total volume to 10 mL. The amount of water needed is:

(10 mL - 0.8 mL) = 9.2 mL

6 0

2 years ago

How many atoms are there in 5.00 mol of sulphur (S)?

Mumz [18]

Answer:

3.01 × 10²⁴ atoms S

General Formulas and Concepts:

Chemistry - Atomic Structure

Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

Step 1: Define

5.00 mol S

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

$5.00 \ mol \ S(\frac{6.022 \cdot 10^{23} \ atoms \ S}{1 \ mol \ S} )$ = 3.011 × 10²⁴ atoms S

Step 4: Check

We are given 3 sig figs. Follow sig fig rules and round.

3.011 × 10²⁴ atoms S ≈ 3.01 × 10²⁴ atoms S

6 0

3 years ago

Look at sample problem 19.10 in the 8th ed Silberberg book. Write the Ksp expression. Find the concentrations of the ions you ne

777dan777 [17]

Answer:

Explanation:

From the information given:

$CaF_2 \to Ca^{2+} + 2F^-$

$Ksp = 3.2 \times 10^{-11}$

no of moles of $Ca^{2+}$ = 0.01 L × 0.0010 mol/L

no of moles of $Ca^{2+}$ = $1 \times 10^{-5} \ mol$

no of moles of $F^-$ = 0.01 L × 0.00010 mol/L

no of moles of $F^-$ = $1 \times 10^{-6}\ mol$

Total volume = 0.02 L

$[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\ \\ \[[Ca^{2+}}] = 0.0005 \ mol/L$

$[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}$

$[F^{-}] = 5 \times 10^{-5} \ mol/L$

$Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}$

Since Q<ksp, then there will no be any precipitation of CaF2

3 0

3 years ago

*WILL GIVE BRAINLIEST IF SOMEONE ANSWERS THIS CORRECTLY ASAP!!!!*

WILL GIVE BRAINLIEST IF SOMEONE ANSWERS THIS CORRECTLY ASAP!!!!