Answer please????? Thank you

Assume the mass of the star is 60. g. Calculate the density. Use correct number of sig figs.

Stels [109]

It’s a bad idea for a few hours but it’s

7 0

3 years ago

7. Halfway through a cross country meet, a

kondor19780726 [428]

Answer:

3m/s

Explanation:

Data obtained from the question include:

Initial speed (s1) = 4 m/s

Final speed (S2) = 7m/s

Change in speed (ΔS)

ΔS = s2 — s1

ΔS = 7 — 4

ΔS = 3m/s

Therefore, the change in speed is 3m/s

7 0

4 years ago

Classify each of these reactions. A single reaction may fit more than one classification. Ba ( ClO 3 ) 2 ⟶ BaCl 2 + 3 O 2 Ba(ClO

Lena [83]

Answer: a) $Ba(ClO_3)_2\rightarrow BaCl_2+3O_2$ : Decomposition

b) $NaNO_2+HCl\rightarrow NaCl+HNO_2$ : double displacement

c) $CaO+CO_2\rightarrow CaCO_3$ : Synthesis (Combination)

d) $ZnSO_4+Mg\rightarrow MgSO_4+Zn$ : redox

Explanation:

Decomposition is a type of chemical reaction in which one reactant gives two or more than two products.

$Ba(ClO_3)_2\rightarrow BaCl_2+3O_2$

A double displacement reaction is one in which exchange of ions take place.

$NaNO_2+HCl\rightarrow NaCl+HNO_2$

Synthesis reaction is a chemical reaction in which two reactants are combining to form one product.

$CaO+CO_2\rightarrow CaCO_3$

Redox reaction is a type of chemical reaction in which oxidation and reduction takes place in one single reaction. The oxidation number of one element increases and the oxidation number of other element decreases.

$ZnSO_4+Mg\rightarrow MgSO_4+Zn$

4 0

3 years ago

I’m very confused on how to solve this

Bond [772]

Recall that density is Mass/Volume. We are given the mL of liquid which is volume so all we need is mass now. We are given the mass of the granulated cylinder both with and without the liquid, so if we subtract them, we can get the mass of the liquid by itself. So, 136.08-105.56= 30.52g. This is the mass of the liquid. We now have all we need to find the density. So, let’s plug these into the density formula. 30.52g/45.4mL= 0.672 g/mL. This is our final answer since the problem requests the answer in g/mL, but be careful, because some problems in the future may ask for g/L requiring unit conversions. Also note that 30.52 was 4 sigfigs and 45.4 was 3 sigfigs, and so dividing them required an answer that was 3 sigfigs as well, hence why the answer is in the thousandths place

4 0

3 years ago

A 59.1g sample of aluminum is put into a calorimeter (see sketch at right) that contains 250.0g of water. The aluminum sample st

Rainbow [258]

Answer:

The specific heat capacity of aluminum according to this experiment is 0.863 J/g°C

Explanation:

Step 1: Data given

Mass of aluminium = 59.1 grams

Mass of water = 250.0 grams

Initial temperature of aluminium = 91.3 °C

Initial temperature of water = 16.0 °C

Final temperature = 19.5 °C

Pressure remains constant

Specific heat capacity of water = 4.186 J/g°C

Step 2: Calculate specific heat of aluminium

Heat lost = heat gained

Qlost = -Q heat

Q = m*c*ΔT

heat aluminium = - heat water

m(aluminium) * c(aluminium) * ΔT(aluminium) = -m(water) * c(water) * ΔT(water)

⇒m(aluminium) = mass of aluminium = 59.1 grams

⇒c(aluminium) = the specific heat of aluminium = TO BE DETERMINED

⇒ΔT = the change in temperature = T2 -T2 = 19.5 - 91.3 = -71.8 °C

⇒ m(water) = 250.0 grams

⇒c(water) = the specific heat of water = 4.186 J/g°C

⇒ΔT = the change in temperature = T2 -T2 = 19.5 - 16.0 = 3.5 °C

59.1 * c(aluminium) * -71.8 °C = 250.0 * 4.186 J/g°C * 3.5 °C

c(aluminium) = 0.863 J/g°C

The specific heat capacity of aluminum according to this experiment is 0.863 J/g°C

3 0

3 years ago