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3 years ago

If a car travels 60 miles in 3 hours, what is the car’s speed?

Chemistry

2 answers:

marysya [2.9K]3 years ago

6 0

Answer = 20 mph !
Hope this helps

Andrews [41]3 years ago

4 0

It's probably about 20 miles per hour because that's REALLY slow

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How many molecules of Ca are found in a sample with 0.2 mols?

DochEvi [55]

Answer:

$\boxed{1.2 \times 10^{23}\text{ atoms}}$

Explanation:

6.023 × 10²³ atoms of Ca are in 1 mol of Ca

$\text{No. of atoms} = \text{0.2 mol} \times \dfrac{6.023 \times 10^{23}\text{atoms }}{\text{1 mol }} = \mathbf{1.2 \times 10^{23}} \textbf{ atoms}}\\\\\text{There are }\boxed{\mathbf{1.2 \times 10^{23}} \textbf{ atoms}} \text{ atoms in 0.20 mol of Ca}$

6 0

2 years ago

In an isolated system:

ipn [44]

Answer:

isolated system (plural isolated systems) (physics) A system that does not interact with its surroundings. Depending on context this may mean that its total energy and/or momentum stay constant.

Explanation:

An isolated system is a thermodynamic system that cannot exchange either energy or matter outside the boundaries of the system. ... The system may be enclosed such that neither energy nor mass may enter or exit.

is there both?

7 0

3 years ago

How do you do empirical formula

mihalych1998 [28]

Answer:

Explanation:

Step 1: Obtain the mass of each element present in grams. Element % = mass in g = m.

Step 2: Determine the number of moles of each type of atom present. ...

Step 3: Divide the number of moles of each element by the smallest number of moles. ...

Step 4: Convert numbers to whole numbers.

7 0

2 years ago

How are electromagnetic waves different from other waves

jenyasd209 [6]

They are different in wavelength.

3 0

3 years ago

Complete combustion of 5.90 g of a hydrocarbon produced 19.2 g of CO2 and 5.89 g of H2O. What is the empirical formula for the h

scZoUnD [109]

$44g \ CO_{2} \ \ \ \rightarrow \ \ 12g \ C\\
19,2g \ CO_{2} \rightarrow \ \ \ x\\\\
x=\frac{19,2g*12g}{44g}\approx5,24g \ \ \ \Rightarrow \ \ n=\frac{5,24g}{12\frac{g}{mol}}\approx0,44mol\\\\\\
18g \ H_{2}O \ \ \ \rightarrow \ \ \ 2g \ H\\
5,89g \ H_{2}O \ \rightarrow \ \ y\\\\
y=\frac{5,89g*2g}{18g}\approx0,65g \ \ \ \ \Rightarrow \ \ n=\frac{0,65g}{1\frac{g}{mol}}=0,65mol\\\\\\
n_{C}:n_{H}=0,44:0,65\approx1:1\\\\
CH$

4 0

2 years ago