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elena-s [515]
3 years ago
13

6th grade science i mark as brainliest​

Chemistry
1 answer:
tino4ka555 [31]3 years ago
3 0

Answer:

its sideways

Explanation:

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Part 1: Fill in the blank<br> Newton’s Second Law: Unbalanced forces cause an object to ______.
blagie [28]

Answer:

and can cause changes in motion. Inertia.

Explanation:

4 0
3 years ago
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A sample of hydrogen gas will behave most like an ideal gas under the conditions of
elena55 [62]

Answer:

The Answer is gonna be C) high pressure and low temperature

8 0
3 years ago
How many calories of heat are required to raise the temperature of 1.00kg of water from 10.2 degrees Celsius to 26.8 degrees Cel
Vika [28.1K]

Answer:

Explanation:

we know that specific heat is the amount of heat required to raise the  temperature of substance by one degree mathmeticaly

Q=mcΔT

ΔT=T2-T1

ΔT=26.8-10.2=16.6

C for water is 4.184

therefore

Q=1.00*4.184*16.6

Q=69.4 j

now we have to covert joule into calorie

1 calorie =4.2 j

x calorie=69.4 j/2

so 69.4 j =34.7 calorie thats why 34.7 calorie heat is required to raise the temperature of water from 10.2 to 26.8 degree celsius

6 0
3 years ago
What mass of glucose would you need (in g) to have 0.8 mol, given that the molar mass of glucose is 180 g mol-1?
Verizon [17]
<h3>Answer: 144 g</h3>

Explanation:

  Mass of glucose = moles × molar mass

∴ Mass of glucose = 0.8 mol × 180 g mol⁻¹

                               =  144 g

∴ the mass of glucose you need to have 0.8 mol of glucose = 144 g

5 0
2 years ago
Given these reactions, X ( s ) + 1 2 O 2 ( g ) ⟶ XO ( s ) Δ H = − 668.5 k J / m o l XCO 3 ( s ) ⟶ XO ( s ) + CO 2 ( g ) Δ H = +
qwelly [4]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

X(s)+\frac{1}{2}O_2(g)+CO_2(g)\rightarrow XCO_3(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) X(s)+\frac{1}{2}O_2(g)\rightarrow XO(s)    \Delta H_1=-668.5kJ

(2) XCO_3(s)\rightarrow XO(s)+CO_2     \Delta H_2=+384.3kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(1\times (-668.5))+(1\times (-384.3))=-1052.8kJ

Hence, the \Delta H^o_{rxn} for the reaction is -1052.8 kJ.

7 0
3 years ago
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